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 1 answerLast reply by: Professor Selhorst-JonesFri May 13, 2016 7:44 PMPost by Raoul Benjamin on May 13, 2016for Example 4:how come it does not work if you write 12 as 4^1.79248 then cancel out the 4's and just solve for t? 2 answersLast reply by: Napolean RichardMon Jan 19, 2015 10:36 AMPost by Napolean Richard on January 6, 2015Dear sir,In this practise quesetion:lnx + ln(x+3) = 2 .Why could not i make both side an e equation exponent to cancel out ln directly?Then the next is x+(x+3)=e^2? 0 answersPost by Saadman Elman on November 11, 2014The concept became very clear now. Thanks!

### Solving Exponential and Logarithmic Equations

• Now that we understand how exponents and logarithms work, how do we solve equations involving them? There are two main ideas that allow us solve such equations: the one-to-one property and the inverse property.
• The one-to-one property points out that exponential and logarithmic functions are both one-to-one, that is, different inputs always produce different outputs.
 ax = ay   ⇔   x = y                      logb x = logb y   ⇔  x=y
If an equation is set up in one of the formats above, we can turn it into something fairly easy to solve.
• The inverse property points out that exponentiation and logarithms are inverses of each other: if they have the same base, they "cancel out".
 loga (ax) = x                             blogb x = x
This means we can make both sides of an equation exponents or take the logarithm of both sides to "cancel out" things that are "in the way" of solving an equation.
• One particularly useful property of logarithms is
 loga xn   =  n ·loga x.
This allows us to "bring down" exponents with any logarithm. Furthermore, we can use logarithm bases that we have on our calculators to make calculation easy (e⇒ ln,  10 ⇒ log).
• Many of the properties we've discussed about exponents and logarithms in previous lessons can be useful in solving exponential or logarithmic equations. If a problem is complicated, try to figure out if you can first simplify it with some of the various properties we've learned.
• While solving these equations, it's important to watch out for extraneous solutions: values that appear over the course of solving, but are not actually solutions. A value might seem like a solution, but actually be outside of the allowed domain. Make sure to check your answers to be sure they do not cause any issues.

### Solving Exponential and Logarithmic Equations

Solve for x.
 472x+9 = 473x−1
• We begin by using the one-to-one property. The only way this expression could be true is if the two exponents are equal to each other. This means we can set them equal in an equation, then solve.
• From the one-to-one property, we have
 2x+9 = 3x−1,
which we can solve for x using basic algebra.
x=10
Solve for t.
 ln(t+17) = ln(6t+2)
• We begin by using the one-to-one property. The only way this expression could be true is if the two quantities that the logarithm acts upon are equal to each other. This means we can set them equal in an equation, then solve.
• From the one-to-one property, we have
 t+17=6t+2,
which we can solve for t using basic algebra.
t=3
Solve for a.
 36a+3 = 6−4a − 6
• Notice that we can't immediately use the one-to-one property because we don't have the same base for the exponent on either side of the equation. However, notice that 36 and 6 are connected through exponentiation, so we can find a way to rewrite one of the sides to match the other side. [We could also approach this problem by taking the log of both sides, bringing down the exponents, then working through it, but that makes things a little more difficult (and could potentially give us an inaccurate answer if we allow any mistakes to creep in).]
• We see that 36=62, so we can substitute that in:
 36a+3 = 6−4a − 6    ⇒     (62)a+3 = 6−4a − 6     ⇒     62(a+3) = 6−4a − 6
• Now that either side has the same base for the exponent, we can make an equation out of the exponents using the one-to-one property:
 2(a+3) = −4a − 6
At this point we can solve for a using basic algebra.
a=−2
Solve for x exactly.
 10−7x+3 = 1000
• The problem here is that our variable is "stuck" in an exponent. We can get it out of the exponent by using an appropriate logarithm. This is called the inverse property:
 loga (au) = u                             blogb v = v
• By taking log of both sides (remember, log with no number is short-hand for log10), we can "free" the exponent:
 10−7x+3 = 1000     ⇒     log( 10−7x+3 ) = log1000     ⇒     −7x + 3 = log1000
• Furthermore, we can use a calculator to find that log1000 = 3 (or we can realize that 1000 = 103 and get the same result), thus we have:
 −7x + 3 = log1000     ⇒     −7x + 3 = 3     ⇒     x=0
• As an alternative method to doing this problem, we could also realize from the beginning that 1000=103, allowing us to make the substitution
 10−7x+3 = 1000     ⇒     10−7x+3 = 103
and then use the one-to-one property to see that
 −7x+3 = 3.
Either way of approaching the problem will work, but it's good to start seeing that logarithms can be used to "cancel out" the bases of exponents we want to work with.
x=0
Solve for x exactly.
 log8 x5 − 2 log8 x3 = log8 x4 − 10
• While we could attempt to put each side as an exponent on a base of 8 to cancel out the log8's, that would get messy. It would be possible, but it would be a little bit of a pain to work with. What we can do instead is try to combine the logarithms as much as possible through the use of logarithm properties and simple algebra before we try to cancel out the log8.
• By the properties of logarithms, we can bring down the exponents-loga xn = n ·loga x:
 log8 x5 − 2 log8 x3 = log8 x4 − 10     ⇒     5log8 x − 6log8 x = 4log8 x − 10
• Notice that we can now simplify things by combining the like terms of log8 x:
 5log8 x − 6log8 x = 4log8 x − 10    ⇒     −1log8 x = 4log8 x − 10
Then, we can manipulate through basic algebra, adding log8 x + 10 to both sides:
 −1log8 x = 4log8 x − 10     ⇒     10 = 5 log8 x
• Continue to simplify a bit more:
 10 = 5 log8 x     ⇒     2 = log8 x
At this point, we can get rid of the log8 by raising both sides up as exponents with a base of 8:
 82 = 8log8 x     ⇒     64 = x
If you're not sure how the above step worked, remember that logarithm and exponentiation of the same base cancel each other out. This is called the inverse property:
 loga (au) = u                             blogb v = v
x=64
Solve for p below. Name the extraneous solution that you find while solving.
 log3 (p2−5) = log3 (5 − 3p)
• Since we have the same log on both sides and nothing else, things are conveniently set up to use the one-to-one property:
 log3 (p2−5) = log3 (5 − 3p)    ⇒     p2−5 = 5−3p
• Since we have a polynomial structure, we solve by getting 0 on one side and factoring:
 p2−5 = 5−3p     ⇒     p2 + 3p −10 = 0     ⇒     (p+5)(p−2) = 0
Thus, solving the polynomial gives us two values for p: −5 and 2.
• However, just because the values are solutions to the polynomial equation does not necessarily mean they are solutions to the logarithmic equation we started with. Remember, for a logarithm to work, it must operate on a number in the set (0, ∞). Thus, we need to check and make sure that plugging in the number won't "break" the logarithm.
 p=−5:       log3 ⎛⎝ (−5)2−5 ⎞⎠ = log3 ⎛⎝ 5 − 3(−5) ⎞⎠ ⇒      log3(20) = log3 (20)

 p=2:       log3 ⎛⎝ 22−5 ⎞⎠ = log3 ⎛⎝ 5 − 3(2) ⎞⎠ ⇒      log3(−1) = log3 (−1)    BAD
Because p=2 causes a negative to appear in the logarithm (thus "breaking" it), we must discard it as a solution. We call solutions that initially appear as possibilities but are later discarded extraneous solutions.
p=−5;        Extraneous solution: p=2 [Note that p=−5 is the only correct solution to the problem, p=2 is just a value that we initially thought would work, but are later forced to throw out as extraneous.]
Solve for x exactly, then also give an approximation to four decimal places.
 4ex = 27
• We can't use the one-to-one property here because we have no corresponding e base on the other side. This means we will have to solve this by using the inverse property to cancel out the e base of ex. We will do this by taking the natural log (ln) of both sides. [Remember, ln is just short-hand for loge.]
• To make it a little easier for us, begin by solving for ex on one side of the equation:
 4ex = 27     ⇒     ex = 27 4
• Now we can take the natural log of both sides so that we can cancel out the e base and get x alone on one side:
 ex = 27 4 ⇒     ln(ex ) = ln ⎛⎝ 27 4 ⎞⎠ ⇒     x = ln ⎛⎝ 27 4 ⎞⎠
At this point, we have solved for x, and we see it is precisely the value of ln([27/4]). We cannot simplify it any more. [This is because the natural log of most numbers is like the square root of most numbers. While there are a very few numbers that will simplify nicely (e, e2, e3, ek, etc.), most numbers cannot be simplified any further. We can write out the precise value using the logarithm, or we can plug it in to a calculator to get a decimal approximation.
• Plugging into a calculator, we find
 x = ln ⎛⎝ 27 4 ⎞⎠ ≈ 1.9095
x = ln( [27/4] ) ≈ 1.9095
Solve for t.
 logt3 + logt2 = 5
• We will want to use the inverse property to cancel out the logarithm, but it will make our work easier to begin by combining the logarithms into a single thing before canceling out. We can do so with the logarithm property loga (M ·N) = loga M + loga N then simplifying:
 logt3 + logt2 = 5     ⇒     log(t3 ·t2) = 5     ⇒     log(t5) = 5
• We can make things even easier by "pulling down" the exponent in the logarithm by using the property loga xn = n ·loga x:
 log(t5) = 5    ⇒     5 logt = 5
Then, using basic algebra to divide by 5 on both sides, we get:
 5 logt = 5     ⇒     logt = 1
• Now that it's in a nice, simple form with the logarithm alone on one side, it is easy to use the inverse property to cancel out the logarithm. If we put both sides as exponents with a base of 10, we can cancel out the logarithm (remember, log is short-hand for log10):
 logt = 1     ⇒     10logt = 101     ⇒     t = 10
t=10
Solve for z precisely, then give an approximation to four decimal places.
 4 · ⎛⎝ 1 3 ⎞⎠ 2z+3 = 8
• Begin by isolating the base/exponent combination:
 4 · ⎛⎝ 1 3 ⎞⎠ 2z+3 = 8     ⇒ ⎛⎝ 1 3 ⎞⎠ 2z+3 = 2
• Next, we want to "get access" to the (2z+3) exponent on the base. At this point, there are two options: first, we could take log[1/3] of both sides. This would cancel out the [1/3] base, but has the downside that very few calculators can tell us what log[1/3] 8 is. [Although we could use the change of base formula later on.] Alternatively, we can use a base that most calculators can easily work with: 10 or e. Either works fine, so let's arbitrarily pick log to work with. While log can't cancel the [1/3] (because it has a base of 10, which is different), we can use one of the logarithm properties that allows us to "pull down" the exponent-loga xn = n ·loga x:
 ⎛⎝ 1 3 ⎞⎠ 2z+3 = 2     ⇒     log ⎡⎣ ⎛⎝ 1 3 ⎞⎠ 2z+3 ⎤⎦ = log2     ⇒     (2z+3) log ⎛⎝ 1 3 ⎞⎠ = log2
Using this method will allow us to easily find the decimal approximation with our calculators at the end.
• Now we can go about solving for z like normal with algebra. Just be careful to treat the logarithms as quantities while working.
(2z+3) log
1

3

= log2     ⇒     2z + 3 = log2

 log 1 3
⇒     z = log2

 2 log 1 3
3

2
• We can't simplify things any further, so now we use a calculator to approximate the value:
z = log2

 2 log 1 3
3

2
≈ −1.8155
• In case you're curious about how we could have done the first method (using log[1/3]), here's a quick run-down. First, cancel using log[1/3]:
 ⎛⎝ 1 3 ⎞⎠ 2z+3 = 2     ⇒     log[1/3] ⎡⎣ ⎛⎝ 1 3 ⎞⎠ 2z+3 ⎤⎦ = log[1/3] 2     ⇒     2z + 3 = log[1/3] 2
Then solve for z:
 2z + 3 = log[1/3] 2     ⇒     2z = log[1/3] 2 − 3     ⇒     z = log[1/3] 2 − 3 2
This is an equivalent way to give z, the only problem is that most calculators can't directly give us a decimal value because of the logarithm base. Use the change of base formula to obtain
log[1/3] 2 = log2

 log 1 3
,
then plug that in when evaluating z to find the decimal approximation.
z = [ log2/(2 log[1/3] )] − [3/2] ≈ −1.8155 [Alternatively, you could give the equivalent answer of z = [(log[1/3] 2 − 3)/2]. Both are correct, just two different ways of getting to the solution. See the steps for more of an explanation.]
Solve for x precisely, then give an approximation to four decimal places.
 lnx + ln(x+3) = 2
• Begin by assembling your logarithms into a single logarithm so that it will be easier to cancel the logarithm out:
 lnx + ln(x+3) = 2    ⇒     ln ⎛⎝ x(x+3) ⎞⎠ = 2     ⇒     ln(x2 + 3x ) = 2
• Using the inverse property, cancel out the logarithm by raising both sides as powers:
 ln(x2 + 3x ) = 2     ⇒     eln(x2 + 3x ) = e2     ⇒     x2 + 3x = e2
• Now we need to solve for x. Notice that e2 is just a number, so we've got a polynomial equation. While e2 is too complicated for us to solve by factoring, we can use the quadratic formula, so begin by getting everything on one side:
 x2 + 3x = e2    ⇒     x2 + 3x − e2 = 0
Now we can set up the quadratic formula. We have a = 1, b=3, and c=−e2:
x = −b ± √ b2−4ac

2a
=     −3 ± √ 32 − (4)(1)(−e2)

2·1
• Simplify:
 −3 ± √ 32 − (4)(1)(−e2)

2·1
=     −3 ± √ 9+4e2

2
We can't simplify any more at this point, so we see we now have two solutions from solving the polynomial equation:
x = −3 + √ 9+4e2

2
≈ 1.6047            x = −3 − √ 9+4e2

2
≈ −4.6047
• However, just because those are the solutions to the polynomial equation does not necessarily mean they will satisfy the original, logarithmic equation. Remember, logarithms can only operate on numbers greater than 0. If we were to plug in the second "solution", we would get
x = −3 − √ 9+4e2

2
: ⇒ ln(−4.6047) + ln(−4.6047 + 3) = 2     BAD.
Since we can't take the logarithm of a negative number, we must throw that out as an extraneous solution.
• While we're at it, it's probably a good idea to check that the other solution does, in fact, work. After a tough problem, it is a good idea to check your answer to make sure there aren't any mistakes We can do this by plugging in our approximate value and using a calculator. Plugging in for the left side, we have
x = −3 + √ 9+4e2

2
: ⇒ ln(1.6047) + ln(1.6047+3)     =     2.000  014…
Since the right-hand side of the original equation is 2, this checks out. [The reason it did not come out precisely to 2 is because of rounding error. We got the value of 1.6047 only after rounding to four decimal places, which introduced a slight error.]
x = [(−3 + √{9+4e2})/2] ≈ 1.6047 [Note: The above is the only solution. If you got two values for x, don't forget to check for extraneous solutions.]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Solving Exponential and Logarithmic Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Introduction 0:05
• One to One Property 1:09
• Exponential
• Logarithmic
• Specific Considerations
• One-to-One Property
• Solving by One-to-One 4:11
• Inverse Property 6:09
• Solving by Inverses 7:25
• Dealing with Equations
• Example of Taking an Exponent or Logarithm of an Equation
• A Useful Property 11:57
• Bring Down Exponents
• Try to Simplify
• Extraneous Solutions 13:45
• Example 1 16:37
• Example 2 19:39
• Example 3 21:37
• Example 4 26:45
• Example 5 29:37

### Transcription: Solving Exponential and Logarithmic Equations

Hi--welcome back to Educator.com.0000

Today, we are going to talk about solving exponential and logarithmic equations.0002

At this point, we have a good understanding of how both exponentiation and logarithms work.0006

However, we haven't seen much about how to solve equations involving them.0010

For example, how do we solve something like e2x - 3 = e5x - 12? Or log(x - 2) = 1?0013

We haven't really talked about how to do that.0023

We briefly touched on it in the final example in the last lesson, but now we are going to really explore it.0024

We will go over two different ways of approaching such equations.0029

First, we will discuss how we can use the 1:1 property, and then we will see how we can apply the inverse property for more complicated equations.0032

Make sure you already understand how exponents and logarithms work.0040

The previous few lessons explain this stuff in detail--explain how all of these things work and their properties.0044

So, make sure you get an understanding of that before you try to get into these equations.0049

While you might be able to understand it, it will make a lot more sense if you already have a grasp0052

of how exponents and logarithms work and what their properties are.0057

Then, you will really be able to throw yourself into the equations.0060

If you don't have a good grasp of that, first make sure that you have watched those lessons previously,0062

because they will really help out for understanding this lesson.0065

All right, notice that the exponential and logarithmic equations are both one-to-one: different inputs imply different outputs.0069

If we put a different value into the function, a different output always will come out.0076

We can see this in the graphs, because both function types pass the horizontal line test.0081

If we cut an exponential function's graph with a horizontal line, it is only going to intersect at a maximum of one point.0085

This means that it is a one-to-one function.0093

If you want a better understanding of one-to-one functions, you want to check out the Inverse Functions lesson.0097

The same with logarithmic functions: if we cut at any place with a horizontal line, it will only intersect one time.0104

Thus, they are one-to-one functions: for any input, there is only one output; for every output that can come out of it, there is a unique input that creates it.0111

As an example, let's consider some specific numbers: if we wanted to see 34 = 3 to the something,0123

what could go into that box to give us 34?0129

Well, if we think about it for a while, we will probably think, "Well, there is only one choice to go into that box."0132

It doesn't make sense for there to be anything else; the only other thing that could be on the other side of the equation is also 34.0137

The only thing that is going to make 34 with a base of 3 is 34.0143

So, the only thing that can go inside of this box is a 4.0148

Similarly, over here with log5(25) = log5(something),0152

the only thing that could possibly be in there...if we think about this for a while, we will see,0157

"Well, the only thing that could make any sense to go inside of that box is a 25."0160

It wouldn't work any other way; there is nothing else that we could take log base 5 of and get the same value as log5(25).0165

So, the only thing that could go in those boxes is the number on the other side of the equation.0174

This is because they are one-to-one--this one-to-one property.0178

Since we have an input of 4 over here, and it gives us an output of 34, 81,0181

we know that the only other input that is going to be able to do that is no other input.0187

This 4 is the only input that can give us 81 as our output; so the same thing must be here--it is one-to-one.0192

If an input gives an output, the only input that can give that output is that original input.0199

There is no other input; it is unique to the output.0205

We call this the one-to-one property; it says that, if the base of a is raised to x, and we have that equal to the same base, a,0211

raised to y, then it must be the case that x equals y.0219

Similarly, if we have logb acting on x, and we know that it is equal to logb acting on y, then we know that x = y.0223

And notice that, in both of these cases, the base has to be the same on each one.0231

We have ax and ay; we have logb and logb--0235

logb(x) and logb(y)--we have the same bases in both of these cases.0241

And that is why we have this one-to-one property working: we know that x has to equal y in both of these situations.0246

All right, with this property in mind, we can now solve equations where we have an exponent or a logarithm of a single base on both sides of the equation.0252

For example, if we have e2x - 3 and e5x - 12, well, since we have the same base on both sides,0259

we know that 2x - 3 has to be the same thing as 5x - 12.0266

Otherwise, we wouldn't have equality.0270

So, we know that 2x - 3 = 5x - 12; and at this point, we go about it just like we are solving a normal algebra equation.0271

Add 12 to both sides; we get 2x + 9 = 5x; subtract 2x from both sides; we get 9 = 3x; divide by 3 on both sides; we get 3 = x.0279

If we want to check it, we can plug in our 3 = x: e2(3) - 3 = e5(3) - 12.0291

So, we have e6 - 3 = e15 - 12, so e3 = e3; and that checks out just fine.0304

The same basic idea over here: log7(x + 5)...and let's move the other log7(2x - 3).0315

We will add log7(2x - 3) to both sides; so it will now appear on the right side.0322

At this point, we know that what is inside of both logs has to be the same thing.0331

They are both log7, so we know that they must be taking logarithms of the same object.0336

x + 5 and 2x - 3 must be the same thing; otherwise, we couldn't have equality.0344

We have x + 5 = 2x - 3 by this one-to-one property.0349

We can add 3 on both sides; we will get 8; so x + 8 = 2x.0355

We can subtract by x on both sides; we will get 8 = x; and there is our answer.0362

And if we wanted to, we could check that one the same way.0367

The inverse property: previously, we have talked about exponentiation and logarithm, if they have the same base, are inverse processes.0370

If they are applied on after another, they cancel each other out.0378

So, if we have natural log on e2, well, since natural log is just the same thing as log base e,0383

and it is operating on something that is base e, to the 2, then the log base e here and the e cancel out, and we are left with just 2.0390

So, sure enough, 2 = 2; that is the idea of what is going on there.0397

The same thing over here: 5 as our base, raised to the log5(125)--they cancel out, and we get 125.0401

That is how this stuff is coming out.0409

If you want a more in-depth exploration of this, check out the previous lesson, Properties of Logarithms,0410

where we will actually prove this stuff and see why it has to be the case.0414

We call this the inverse property: so loga(ax) is equal to x.0418

And blogb(x) = x; so we have this cancellation.0424

If we end up having the same base like this in log base, and then exponent base, or in exponent base and then log base,0429

we will cancel out, and we will just get the thing that is at the end.0437

So, in this case, that is x here or x here that we end up getting out of it.0440

Solving by inverses: with an equation or inequality, we can do algebra.0446

Now, algebra is just applying the same thing to both sides; we are doing the same operation--whatever it is.0451

When we first learned to do algebra, we just used simple arithmetic (things like addition, subtraction, multiplication, or division).0458

what we just saw in the previous thing when we were solving after we used these special properties.0466

But as we learned more in our work in algebra, we realized that we could do more than just apply simple operations on both sides.0470

We could even do more complicated things, like squaring both sides, or taking the square root.0477

We realized that, as long as we are doing the same action to the left side and the right side--0482

we are doing the same thing to two things that are equal--0486

we know that it is going to remain being equal, even after the action goes through.0489

So, by that same idea, we can also make both sides of an equation exponents, or take the logarithm of both sides,0493

because we are doing the same action to both sides--the equality is still based in it.0499

If we have wavy stuff equals loopy stuff, then we know that a, as a base for wavy stuff, is going to be equal to a as a base for loopy stuff,0504

because a to the something-on-the-left equals a to the something-on-the-right, and we were told that left side and right side are the same.0515

So, it must be the same still, even when they are working as bases for something else,0521

or when something is a base underneath them and they are now exponents.0525

A similar idea: if we were told that wavy equals loopy, then we know that logb(wavy) must be the same thing as logb(loopy).0528

We are applying the same action, whether it is turning them into exponents on some base,0536

or we are taking the logarithm of some base of both sides.0540

We are doing the same action to both sides, so we have this equality still holding.0543

Combining this idea with the inverse property allows us to get rid of exponent bases or logarithms that are in the way of solving an equation.0547

For example, if we have log(x - 2) = 1 (remember, if it is just log, then that is the common log, so it is log base 10 of x - 2 equals 1);0556

what we can do is come along and raise both sides with a 10 underneath them.0566

So, we are not raising to a power, like squaring them; we are actually causing this exponent base to erupt underneath them.0572

We have 10 to the log10(x - 2); and that is going to be equal to...10 has erupted underneath the 1.0579

All right, at this point, we have the inverse property; we are solving by inverses, so we have 10 and log10.0589

So, these cancel out, and the x - 2 just drops down; and we get x - 2 =...10 to the 1 is just 10.0595

So now, we just solve it normally: x = 12--there is our answer.0603

A similar idea is going on over here: 32x = 7--well, let's get rid of that base of 3; that is getting in our way.0608

So, we will take log3(32x), and that is going to be equal to log3 of what is on the right side, so log3(7).0615

log3(32x)...those will end up canceling out, and the 2x will just drop down;0627

so we will have plain 2x = log3(7).0634

Now, if we want to figure this out with a calculator...0639

log3(7) is still correct, but if we want figure it out with a calculator--if we actually want a decimal version--0641

we will have to turn it through change of base; let's take natural log--I like natural log.0647

ln(7)/ln(3): remember the change of base formula that we talked about in the previous lesson.0653

x =...now we are dividing by 2 on both sides, so it will show up in the denominator: 2 times the natural log of 3,0659

which will end up working out to approximately 0.8856.0666

Now, notice: 0.8856 is approximate; it is not the exact answer.0672

This is actually the exact answer; once we have calculated through with ln(7) and ln(3), we end up getting something0677

that is very, very close, but it is no longer precise, because we are having to cut off some of the decimal places.0684

And if we wanted to, we also could have used any other base.0689

We could have used log10(7)/log10(3)--a common log there.0691

We would have had log(7)/2log(3), which would end up coming out to be the same thing when we used a calculator on it.0699

And this would also be just equally as correct an answer.0713

A useful property: one particularly useful property of logarithms is this ability to bring down things.0718

If we have loga(xn), then that is the same thing as if the n had been in the front, if it was n times loga(x).0724

So, we can bring down exponents with any logarithm.0731

This means that we can use logarithm bases that we have on our calculators.0735

That might be convenient for us sometimes.0738

So remember: e is the same thing as natural log; 10 is the same thing as log without a number on it.0740

So, if we want, we could just start by taking the natural log of 32x = the natural log of 7.0745

At this point, we can bring down the 2x; it comes down in front by this property up here.0752

So, we have 2xln(3) = ln(7), so 2x = ln(7)/ln(3), or x = ln(7)/2ln(3), which is the exact same thing that we just had on the previous slide.0758

The thing to notice here is that we have two different ways of doing this.0780

We could go about it by using the change of base, or we could go about it by just using this property where we can bring down exponents.0783

Sometimes it will be more useful to use the bringing down the exponent property.0789

Sometimes it will be more useful to do the change of base.0792

They will both end up working out; it is just that sometimes one will be a little bit more work than the other.0794

And you will get a feel for which one you want to use as you work on these things.0798

Many of the properties we have discussed about exponents and logarithms can be useful in solving exponential or logarithmic equations.0802

If the problem is complicated, try to figure out if you can first simplify it with some of the various properties we have learned.0808

We have learned a lot of properties by this point, about how exponentiation works and about how logarithms work.0813

And sometimes, by combining things or breaking things apart, you will make the problem easier to do.0818

And we will see some of that in the examples later on.0823

Extraneous solutions: while solving these equations, it is important to watch out for extraneous solutions.0826

An extraneous solution is a value that appears over the course of solving, but isn't actually a solution--0832

that, if we were to try to use it, would just fail or cause our equation to break apart or not work or not be defined for some reason.0837

The easiest way to see how this works is to just see an example.0844

So, let's look at this: we have the natural log of x2 - 2 equals natural log of x.0847

By the one-to-one property, we see that x2 - 2 has to be equal to x.0851

Alternatively, if we wanted, we could put e's underneath it, and just cancel out both of them.0855

Inverse or one-to-one property--both end up working as ways to look at this.0858

x2 - 2 = x: at this point, it looks like the polynomials we are used to solving from that section.0862

We move it over: x2 - x - 2 = 0; so we can factor that.0870

We get (x - 2)(x + 1) = 0; we solve both of those, so x - 2 = 0; x + 1 = 0; we get x = 2 and x = -1.0875

Now, if we go back and try to work this out, if we plug in x = 2, things are pretty reasonable.0890

We get ln(22 - 2) = ln(2); so this is ln(4 - 2) = ln(2); and then, ln(2) = ln(2).0895

That is perfectly reasonable; but if we try x = -1, we will see some problems very quickly.0910

ln((-1)2 - 2) = ln(-1); and as soon as we see this right here, we get suspicious,0917

because what is the problem here? You can never take the logarithm of a negative number.0928

So, as soon as we see ln of a negative number inside, we know that this is not possible.0933

We can't take the natural log of -1; we can't have logs of negative numbers at any point showing up.0938

This is an extraneous solution--it is something that appeared over the course of solving, because we turned it into this quadratic form.0943

And in the quadratic form, it was a solution; but up here, in the original form that we have, it fails to be a solution.0952

It can't be a solution, because we end up having this logarithm of a negative number; so we knock it out--it is an extraneous solution.0960

And our only answer is x = 2.0967

So, it seems at first as though we have two answers, because we are solving a quadratic.0970

But as we work our way through the quadratic, we realize that if we were to actually plug this in0974

and try it out to see if it works as a solution, it would cause the whole thing to blow apart.0978

So, we end up seeing that it can't actually be used as a solution; so it is called an extraneous solution--0982

something that appears over the course of solving, but can't actually be used as a solution.0987

So, we have only one out of this, even though at first it seemed as though there would be two.0993

All right, we are ready for some examples now.0998

The first one: Solve for x exactly if we have 3/7x + 2 and (49/9)x - 2.1001

So, we look at this, and we think, "Well, we could take logarithms of both sides; we could bring down our exponents."1008

But things are going to get pretty messy; we will have to actually figure out what the logs are of 49/9 and 3/7,1014

and we will have to work out a bunch of numbers.1021

It is going to get really, really ugly: we could work it out that way, but we would end up having approximations,1023

because we would be taking the logs of these numbers, and they would come out to be decimals.1027

So, that won't end up working in the end.1030

But if we look at it, we might realize that 49/9...there is a connection to 3/7.1032

Well, if we want, we could rewrite this as 9/49 to the -1.1039

And then, we might realize that 9 is just 32; 49 is just 72; it is still all to the -1.1045

We can pull out the 2's, and we have (3/7)-2.1052

That is what we started with on the left side; so we can use that one-to-one property.1057

So, we take this fact here; we can swap them out; so the same thing is still on the left side, (3/7)x + 2,1062

is equal to...we swap out 49/9 for (3/7)-2, so we have ((3/7)-2)x - 2.1070

Well, that is going to be equal to (3/7)...we can bring that -2 out by the property of exponents.1083

It will multiply everything that is already out there, so we have to have that and the quantity as well.1090

So, we have (3/7)x + 2; at this point, we can use the one-to-one property, because we have the same base here and here.1094

We have (x + 2) = -2(x - 2); x + 2 = -2x + 4; add 2x to both sides; we get 3x; subtract by 2 on both sides, and we get 2,1102

and we get 3...sorry, we now divide by 3; we don't divide by 2--that would be going the wrong way.1116

Divide by 3 on both sides; we get x = 2/3; and there is our answer for x.1122

And if we wanted to, we could plug that in and check--use our calculator and end up working it out.1128

We would get decimal answers that would end up being the same thing; we would see that that ended up working.1131

So, you can check this by using a calculator if you want to; you could do a check, and you would have (3/7)2/3 + 2 = (49/9)2/3 - 2.1135

You would have to use a calculator to work this out, but if you did, you would get decimal answers that were very, very, very close--1151

actually, they should be exactly the same, because we solved for x exactly.1157

The only problem might be if your calculator has just a little bit of sloppiness in it.1160

But they should get decimal answers that are well within 5 or 10 decimal digits of each other.1164

And so, you will end up seeing that it checks out when you use your calculator.1169

Or you could also just work through each of these, and then use the properties of exponents.1173

And you could see that it is exactly the same thing--there are two ways to do it.1177

Solve for a exactly if we have log(a3) - log(a2) = 2 - log(a).1181

All right, let's use the properties of logarithms to bring some things together and simplify things a bit.1187

Remember: we have subtraction here--subtraction of logarithms is the same thing as division inside of the logarithm.1191

So, we have a3/a2 = 2 - log(a) (not base a).1198

Now, the only issue we would have is...what if a was equal to 0?1206

Well, if a was equal to 0, we would already have problems, because we would be taking the log of 0.1210

So, we don't have to worry about that; so we know that a is not equal to 0, so we are good there.1214

We know that a is not equal to 0, so we can do this cancellation.1219

We don't have to worry about that..1223

log(a3/a2)...well, we will get just log(a), because a3/a2 will cancel two of the a's on top.1224

We will be left with just one.1231

2 - log(a): at this point, we can add log(a) to both sides, so we will get log(a).1233

And now, there are 2 of them, because we added, and they are of the same type; so log(a) + log(a) is 2log(a) = 2.1238

At this point, we can divide by 2 on both sides, and we get log(a) = 1.1246

We want to know what that is: well, remember, log is just common logarithm, so it is base 10.1251

So, we can raise both sides to the 10; so that cancels this out, and we have a = 101, which means a = 10.1256

Great; if we wanted to, we can check this; so as a check, we have log(103) - log(102 = 2 - log(10).1266

log(103) comes out to be 3, because it is base 10, so what do you have to raise 10 to, to get 103?1278

Well, you have to raise it to 3; the same idea is over here--log(102) is just 2, equals 2 - log(10).1285

log base 10 of 10 is just going to be 1; 1 = 1; great--that checks out.1292

All right, the third example: Solve for x to four decimal places: 5 x + 4 = 112x.1297

So, we could write this log5 acting on 5x + 4, and then log5 on the other side, as well, acting on 112x.1304

So, since we have log5 and exponential base 5, they cancel out, and we have x + 4 on the left side.1318

On the right side, we see that we have 2x raised to an exponent; so if we want, we can bring that out to the front.1327

We have equals 2x times, and then our remaining log5(11).1333

All right, so at this point, we can divide by 2x on both sides, because we want to try to get our x's together.1341

Or actually, better yet, we can subtract x on this side, and we will get 4 = 2x(log5(11)) - x.1348

Now, at this point, we can see that there is an x here, and there is an x here; we can pull out the x's.1369

And we will get 4; pull out the x's; x times 2 times log5(11) - 1.1373

We can divide this over, so we have x =...dividing over, we have 4 divided by what we are dividing over, 2log5(11) - 1.1385

We can use the change of base formula, x = 4 over (because we probably wouldn't be able to use a calculator,1399

and lots of calculators can't do log base 5) 2 times log5(11)...let's go with natural log,1406

just because I like natural log; so ln(11)/ln(5) - 1.1413

Now, that one looks kind of ugly, and it is; it is going to take some work through a calculator.1421

But you work it through with a calculator, and you will get that that is approximately equal to 2.0204 once you round it down.1424

So, there you are: another way to have done this would have been to take the natural log of both sides.1435

We could have taken ln(5x + 4) = ln(112x), and this would be true with any base.1440

We could be doing this with the common log, as well, if we wanted to.1449

So, we can bring down our exponents; we will get (x + 4), remember, as a quantity, because it is the whole exponent,1451

times the natural log of 5, equals 2x times the natural log of 11.1457

At this point, we could move the natural log of 5 over, and we would have (x + 4) = 2x[ln(11)/ln(5)].1462

It is over the whole thing, but we can also just compact it into that one thing.1476

And then, if we want, we could move the x over as well, and we would have 4 = 2x[ln(11)/ln(5)] - x.1479

Now, we do the same trick and pull out our x's: we get 4 = x times 2 times ln(11)/ln(5) - 1.1490

We divide that over, and we get 4 over 2 times ln(11)/ln(5) - 1 = x, which is the exact same thing that we had when we did it by using log base 5.1501

So really, it is just a question of if we are prolonging the change of base or causing the change of base to happen as we take a log in a different base.1515

So, there are two different ways to do it.1523

If you don't end up realizing this x trick, where the fact that we have x here and we have x here1525

means that we can pull them both out to the front and get x at the front, you can also just work this out1530

through a bit more arithmetic and working at it, moving things around.1535

You can eventually get it; you could also just do it by evaluating what is log5(11).1540

You could figure out what log5(11) is, get that down to 8 decimal places, and then multiply that be 2x.1546

And you could work it out--just work with a whole bunch of decimal places for a while and solve for what x is,1552

and then just cut it down to four decimal places--that is another way to do it.1557

There are a bunch of different ways that you can approach problems like this.1560

Just remember all of the properties that we have talked about, and just work through it and pay attention to what you are doing.1563

And then, at the very end, check your work--do a quick check.1568

You can do a check that 52.0204 + 4 = 112(2.0204).1571

And in fact, you will end up finding out that they come out to be very, very close.1583

They will end up being different after the fifth or sixth digit, because there was a little bit of rounding error.1587

We did, after all, only round to four decimal places, and it just keeps going forever.1592

But beyond that little bit of rounding error, after the fifth or sixth digit, you will end up being really, really close to being exactly the same.1596

So, you will see that it checks out--that you do have the right answer.1603

The fourth example: Solve for t exactly--we are back to solving for t exactly,1606

so we can't really use these numerical methods that we have been working out with a calculator.1611

We need to figure out something clever here.1615

The problem here is that we have 2t and t; it is not just the same thing happening up there.1617

And then, we also have this confusion from the 12.1623

If the 12 weren't there, we could just move them over, and we could use the one-to-one property.1625

But we have this problem where we have this 42t - 4t = 12...well, let's try moving things around and see if this looks like something we are used to.1629

4t - 12 = 0; at this point, we might have a moment of understanding.1639

We might realize that this looks a lot like a polynomial--like a quadratic polynomial that we are trying to solve.1645

We realize that this is squared; this is to the one; and this is to the nothing--this is the constant.1654

So, we realize that 4t is kind of like the x that we are used to; so let's say 4t = u.1659

We will do this as a u substitution, where we will replace something complicated with something simple, this nice u.1665

If it is 42t, it is now going to be u2,1671

because u2 would be (4t)2, which is the same thing as 42t.1674

Minus u minus 12 equals 0--at this point, it is really easy to solve.1680

We see (u - 4) (u + 3) = 0; great--that makes perfect sense.1685

u times u is u2, plus 3u minus 4u--that is our -u; -4 times +3 is -12; it checks out.1694

So, we solve each one of these independently: u - 4 = 0, so we get u = 4.1701

Now, what is it?--because we are actually solving for the t; we are not solving for u.1706

So, we replace it with 4t = u; so we have 4t = 4, and there is only one thing that is going to end up giving us that.1711

It must be 41 there, so we have t = 1.1718

t = 1 is one of our answers; u + 3 = 0, so we have u = -3; we swap out for 4t...equals -3...1722

and at this point, we realize that that is madness; there is no possible way that 4 to the t can equal a negative number.1734

There is no real number that we can raise 4 to that is going to give us a negative number.1740

So, this way lies madness, so we knock it out; there are no possible answers there.1746

So, our only answer is t = 1; it seemed like we were going to get this answer, but then we realized that that is not possible.1750

If you want to check it out, we plug it in as a check: 42(1) - 41 = 12; 42 - 4 = 12; 16 - 4 = 12.1758

Indeed, that is true; so it checks out.1774

The final example: Solve for x exactly; then give an approximation.1777

We have the x that we are looking for; it is in the denominator.1782

We don't like things stuck in the denominator, so let's multiply it on both sides; we will multiply it by (ln(2x) - 3) on both sides.1785

So, we get 4 = 2 times (ln(2x) - 3) on both sides.1792

We can distribute the 2: 4 = 2 times (ln(2x) - 3); the thing we want to figure out is...it is like we are solving for ln(2x).1799

And then later, we will crack the log; but for now, let's just figure out solving for ln(2x), and then we can solve the log.1810

So, we see the -3 here; we add 3 to both sides, so we get 7 = 2 times...oops, I made a mistake!1817

An important thing to notice: it is 2 times (ln(2x) - 3), so it is 2 times -3; it is not minus 3; it is minus 6.1826

All right, so I have to always be careful with distribution, because it can catch anyone, including me and including your teachers.1836

You always have to watch out for distribution.1842

Add 6 to both sides; we get 10 = 2(ln(2x)); we divide by 2 on both sides; we get 5 = ln(2x).1844

We now raise both sides to the e; so e5, eln(2x).1855

Since these are the same base, natural log and e, they cancel out, and we have e5 = 2x.1861

So, we have that e5/2 = x; that is our exact answer, e5/2.1869

Now, e is this complicated irrational number; we can't really turn it into...it is not exactly a decimal number.1877

But sometimes you want to have a decimal approximation, because that makes it easier for us to work with things.1884

So, e5/2 is the exact answer; that is what it is precisely.1889

But if we want an approximate answer, e5/2 ends up being approximately 74.207; great.1893

Now, if we want to do a quick check, we can do a numerical approximation, where we try plugging it in using a calculator.1906

So, we would have 4 over the natural log of 2, times...replace our x, which we know is approximately 74.207, minus 3.1911

You work that through with a calculator, and you end up getting approximately 1.99999, and then it ends up changing after that.1923

But that is really, really close; so we know that this checks out numerically.1932

We have gotten that close to being exactly 2; so we see that numerically (because remember:1935

there is some rounding error when we get an approximation), with that little bit of rounding error, we are still extremely close.1940

So, we know that that is a good answer.1947

If we wanted to, we could also work it out and show that it ends up being a precise answer, as well--1949

that e5/2 is going to be equal to precisely what x has to be.1954

We can check by plugging that in: 4 over the natural log of 2, times e5 divided by 2, minus 3.1958

So, 2 times...dividing by 2...they cancel out; so we have 4 over the natural log of e5, minus 3.1970

e5 is going to end up being...4/ln(e5) has to be 5, because what do we raise e to1981

(since the natural log is base e) to get e5? Of course, we raise it to a 5 exponent.1991

5 - 3...we could also think about the fact that natural log is log base e, and we have an exponent base e, so they cancel each other out.1997

4 over 5 minus 3 becomes 2, which equals 2; that checks out.2006

So, we can do it either numerically or precisely, and see that it worked in either case.2011

All right, great; now I have a really good understanding of even probably the most complex kind2017

of logarithmic and exponential equations that will be thrown at you at this point in math.2021

So, with this sort of knowledge, you can go and do all kinds of problems, from the easy to the hard ones.2025

And you will be able to solve them if you follow these things.2030

Remember: be careful--mistakes happen; you even saw one happen to me.2032

So, they happen to everybody; it is really useful to do these checks.2036

By checking your work, you can make sure that you didn't make a mistake.2039

And if you see that something went wrong in the check, you can go through and carefully analyze your work and figure out where things went wrong.2042

All right, we will see you at Educator.com later--goodbye!2048