INSTRUCTORS Carleen Eaton Grant Fraser

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• ## Related Books

### Logarithms and Logarithmic Functions

• Remember that a logarithm is just an exponent.
• Understand that the log function and the exponential function are inverses of each other.
• Use this to solve problems.
• Solve logarithmic equations with the same base by equating the expressions whose logarithms have been equated.
• Solve logarithmic inequalities with the same base by applying the same inequality to the expressions whose logarithms have been compared.
• In solving logarithmic equations or inequalities, always check for extraneous solutions – values which result in taking the logarithm of a non-positive value in the original equation or inequality. Exclude such values from the solution set.

### Logarithms and Logarithmic Functions

Solve log4(x4 − 17) = 3
• Recall that logbx = y is the same as by = x
• b = 4
• y = 3
• x = x4 − 17
• Rewrite into exponential form.
• 43 = x4 − 17
• 64 = x4 − 17
• Solve
• x4 = 81
x = 3
Solve log4(x4 − 192) = 3
• Recall that logbx = y is the same as by = x
• b = 4
• y = 3
• x = x4 − 192
• Rewrite into exponential form.
• 43 = x4 − 192
• 64 = x4 − 192
• Solve
• x4 = 256
x = 4
Solve log4(x4 − 369) = 4
• Recall that logbx = y is the same as by = x
• b = 4
• y = 4
• x = x4 − 369
• Rewrite into exponential form.
• 44 = x4 − 369
• 256 = x4 − 369
• Solve
• x4 = 625
x = 5
Solve log2(x3 − 713) = 4
• Recall that logbx = y is the same as by = x
• b = 2
• y = 4
• x = x3 − 713
• Rewrite into exponential form.
• 24 = x3 − 713
• 16 = x3 − 713
• Solve
• x3 = 729
x = 9
Solve log5(x2 − 12) = log5(x)
• Notice that since the bases are the same, you can use the following property
• logbx = logby; then x = y
• log5(x2 − 12) = log5(x)
• x2 − 12 = x
• x2 − x − 12 = 0
• Factor
• x2 − x − 12 = (x − )(x + )
• x2 − x − 12 = (x − 4)(x + 3) = 0
• Solve using the Zero Product Property
• x − 4 = 0;x + 3 = 0
• x = 4;x = − 3
• Check Solutions
•  x=4 x=-3 log5(x2 − 12) = log5(x) log5(x2 − 12) = log5(x) log5(42 − 12) = log5(4) log5(( − 3)2 − 12) = log5( − 3) log5(4) = log5(4) log5( − 3) = log5( − 3) x = 4 is valid Not valid, you cannot have negative logarithms.
x = 4
Solve log5(x2 + 4) = log5( − 5x)
• Notice that since the bases are the same, you can use the following property
• logbx = logby; then x = y
• log5(x2 + 4) = log5( − 5x)
• x2 + 4 = − 5x
• x2 + 5x + 4 = 0
• Factor
• x2 + 5x + 4 = (x + )(x + )
• x2 + 5x + 4 = (x + 1)(x + 4) = 0
• Solve using the Zero Product Property
• x + 1 = 0;x + 4 = 0
• x = − 1;x = − 4
• Check Solutions
•  x=-1 x=-4 log5(x2 + 4) = log5( − 5x) log5(x2 + 4) = log5( − 5x) log5(x2 + 4) = log5( − 5x) log5(( − 1)2 + 4) = log5( − 5( − 1)) log5(( − 4)2 + 4) = log5( − 5( − 4)) log5(5) = log5(5) log5(20) = log5(20) x=-1 is valid x=-4 is valid
x = − 1; x = − 4
Solve log3(5x − 3) < 3
• Recall that logbx = y can be written in exponential form as by = x.
• Don't forget that there's always a restriction when working with logs, namely
• logbx < y; then 0 < x < by
• Solve
• 0 < x < by
• 0 < 5x − 3 < 33
• 0 < 5x − 3 < 27
• 3 < 5x < 30
[3/5] < x < 6
Solve log12(9x − 18) < 2
• Recall that logbx = y can be written in exponential form as by = x.
• Don't forget that there's always a restriction when working with logs, namely
• logbx < y; then 0 < x < by
• Solve
• 0 < x < by
• 0 < 9x − 18 < 122
• 0 < 9x − 18 < 144
• 18 < 9x < 162
2 < x < 18
Solve log7(9x − 18) < 2
• Recall that logbx = y can be written in exponential form as by = x.
• Don't forget that there's always a restriction when working with logs, namely
• logbx < y; then 0 < x < by
• Solve
• 0 < x < by
• 0 < 10x + 9 < 72
• 0 < 10x + 9 < 49
• − 9 < 10x < 40
− [9/10] < x < 4
Solve log3(8 + 3x) < log3(x2 − 2)
• Since the base of the exponents is the same, we can take what is inside the parenthesis outside
• 8 + 3x < x2 − 2
• Move everything to one side of the inequality
• 0 < x2 − 3x − 10x
• Factor
• 0 = (x − 5)(x + 2)
• Solve using the Zero Product Property
•  x − 5 = 0
 x + 2 = 0
•  x = 5
 x = − 2
• In this case, x <− 2 and x > 5 Now check restrictions by the logs
• log3(8 + 3x) < log3(x2 − 2)
• 8 + 3x > 0 3x >− 8 x >− [8/3]
• x2 − 2 > 0 x2> 2 x >√2
• Notice how there's two situations here. while it is true that x can be less than − 2, the restriction
• in the log forces this value to be grater than − [8/3]. Also, the second restriction, that x be greater than √2
• is taken care by the fact that x has to be greater than 5. Therefore, putting this together the solution is
− [8/3] < x <− 2 and x > 5

*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.

### Logarithms and Logarithmic Functions

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
• What are Logarithms? 0:08
• Restrictions
• Written Form
• Logarithms are Exponents
• Example: Logarithms
• Logarithmic Functions 5:14
• Same Restrictions
• Inverses
• Example: Logarithmic Function
• Graph of the Logarithmic Function 9:20
• Example: Using Table
• Properties 15:09
• Continuous and One to One
• Domain
• Range
• Y-Axis is Asymptote
• X Intercept
• Inverse Property 16:57
• Compositions of Functions
• Equations 18:30
• Example: Logarithmic Equation
• Inequalities 20:36
• Properties
• Example: Logarithmic Inequality
• Equations with Logarithms on Both Sides 24:43
• Property
• Example: Both Sides
• Inequalities with Logarithms on Both Sides 26:52
• Property
• Example: Both Sides
• Example 1: Solve Log Equation 31:52
• Example 2: Solve Log Equation 33:53
• Example 3: Solve Log Equation 36:15

### Transcription: Logarithms and Logarithmic Functions

Welcome to Educator.com.0000

Today, we are going to talk about logarithms and logarithmic functions, beginning with the definition.0003

What are logarithms? First, the restrictions: x cannot be 0, and b needs to be a positive number, but not equal to 1.0011

So, the logarithm of x to the base b is written as follows: logb(x).0021

Let's start out with logb(x) = y: in this case, this log is defined to be the exponent y, which would satisfy this equation.0029

Over here, I am talking about the same base: the base b here is the base b right here.0042

We worked with exponents already; and logarithms are actually just exponents.0050

When you use logarithmic notation, you are just writing an exponential expression in a different way.0055

The logb(x) = y is defined as the exponent y that, when you raise b to that power, will give you x back.0062

So, you need to be able to get comfortable with going back and forth between the logarithmic notation and the exponential expression.0076

These two statements are actually inverses of each other; and we will talk more about that relationship in a little while.0084

But we are just starting out now, to get used to the idea of converting back and forth.0090

And the reason that you need to be able to convert back and forth and understand the relationship between these two0094

Given log4(16) = 2, you can rewrite that into the exponential expression.0110

At first, you might need to just stop and analyze the various components.0118

The base is 4; here, x is 16, and y is 2; so I am going to rewrite that: the base remains the same:0123

4 to some power y (here it is 2) equals 16; and we know that that is true--that 42 is 16.0132

You can also move from the exponential equation into the logarithmic equation.0140

Looking at an example where you are starting out with 23 = 8:0145

the base here is 2; y = 3 (that is the power you are raising the base to); and x = 8.0152

Now, I am going to write this as a logarithmic equation: the base is still 2; x is 8; and y is 3; so log2(8) = 3.0164

We can also use this for slightly more complicated situations, such as log5(1/125) = -3.0183

But it is the same idea, because I still have the base, 5, and I know that this is y, so 5-3 equals 1/125.0194

And I know that that is true, because 5-3 would be the same as 1/53, which is 1/125; so that holds up.0205

I can use this to evaluate logarithmic expressions when I am trying to solve.0221

Now, here I gave you all of the pieces; you already had all of the numbers, and it was just rewriting them a different way.0226

Let's look at a situation where we are actually trying to find a value.0232

You are given log3(81) = y.0236

I have the logarithmic expression; but I can solve this more easily if I rewrite it as an exponential equation.0246

I have this base, 3; and what this is saying is that a logarithmic expression is defined to be the exponent y satisfying this relationship.0254

So, if I take the base, and I raise it to the y power, I am going to get 81.0266

I just need to figure out what I would need to raise 3 to, to get 81.0276

3 squared is 9; 3 cubed is 27; 3 to the fourth is 81; therefore, y equals 4, because 34 is 81.0282

This shows you how being able to convert between the logarithmic equation and the exponential equation can help you to solve either one of them.0296

OK, looking on at logarithmic functions: a logarithmic function is a function of the form f(x) = logb(x).0310

We just introduced this idea of logarithms; now we are talking about logarithmic functions, where b is greater than 0 but not equal to 1.0321

And you will call that these are the same restrictions that we had when talking about exponential equations.0330

And this would be for the same reasons as discussed in that lecture, because again, logarithms are simply exponents.0337

As I mentioned in a previous slide, these two are inverses; so f(x) = logb(x)0347

is the inverse of the exponential equation, expression, or function g(x) = b(x).0355

So, f(x) = logb(x) is the inverse of g(x) = bx.0363

And this is an important relationship, because it helps us to solve the equations that we will be working with shortly.0374

Let's just take an example to make this more concrete.0380

Let's let f(x) equal log2(x): the inverse of that would then be g(x) = that same base, 2, raised to the x power.0385

Let's look at some values for f(x): if x is 1, what is y?0400

Well, think about what this is saying: this is saying log2(x) equals some value of y.0408

Rewriting that as a related exponential expression, this is telling me that, when I take 2 and I raise it to the y power, I am going to get x back.0415

What I want to figure out here is: 2 to some power, y, equals 1; what would y have to be?0425

y would have to be 0; therefore, 20 = 1 satisfies this: x is 1 when y is 0.0432

2 to some power y equals 2; what would y have to be? It would have to be 1.0444

How about 4? 2 to some power y equals 4; 22 = 4; therefore y = 2.0451

Let's let x be 8: 2 to some power y equals 8: well, 23 is 8; therefore, y = 3.0461

All right, so that is f(x); now let's look at g(x).0469

If f(x)...this is f(x), but here I am calling it y; now let's look at g(x)...and g(x) are inverses,0475

then what I am going to expect is that the domain of f(x) is going to be the range of g(x),0484

and the range of f(x) is going to be the domain of g(x).0490

So, I am going to go ahead and take these values right here that are the range of f(x);0493

and I am going to use them as the domain of g(x), and see if I get these values back.0500

So, when x is 0, y is 2 to the 0 power, or 1; when x is 1, y is 2; when x is 2, y is 22, is 4.0509

When x is 3, y is 23: it is 8; and I look, and the domain here is equal to the range; the range here is equal to the domain.0529

Now, that doesn't prove anything: it just shows that this one example holds up.0540

But our finding was, as expected, that if f(x) and g(x) are inverses, I do expect the domain of one to be the range of the other,0544

and the range of this one to be the domain of that one.0552

Looking at the graphs of these functions, we can use a table of values to graph a logarithmic function,0560

just as we have used tables of values to evaluate functions earlier in the course, including exponential functions.0566

So, we already started a table for f(x) = log2(x), and for g(x) = 2x, the inverse.0572

So, let's keep going with those, but add on some values.0584

Recall that I said, if f(x) = log2(x), then if I take 2y, I am going to get x.0591

I am rewriting this in exponential form to make it easier for me to find y, because it is difficult to find f(x), or y, when it is in this form.0600

Recall that I said that, when x is 1, then what this is saying is that 2y = 1.0610

And I said that y must be 0; when x is 2, 2y = 2, so y must be 1.0617

And I went on and did a couple of other values, and I did 8: I said 2y = 8; therefore, y had to equal 3.0628

Now, let's add some values: let's add some fractions to get a better idea of what this graph is doing as x becomes small, as it gets close to 0.0639

When x is 1/2, that is telling me 2y = 1/2.0651

The way I would get that is if I took 2-1; I would get 1/2; therefore, y is -1.0657

1/4: 2y = 1/4--if I took 2-2, I would get 1/4, so y is -2.0668

1/8: using that same logic, 23 is 8, but 2-3 would be 1/8, so I am going to make that -3.0678

And then, I am going to graph these values.0688

When x is 1, y is 0; when x is 2, y is 1; when x is 4, y is 2; and when x is way out here at 8, y is 3.0691

So, the general shape is just going up like this.0705

Small values: when x is getting smaller and smaller, what is going to happen with y as x is approaching 0?0709

When x is 1/2, y is -1; when x is 1/4, y is -2; when x is 1/8, y is -3.0719

And what I can see is happening here is that the y-axis is a vertical asymptote.0730

And we talked earlier about the graphs of exponential functions: we saw that the x-axis formed a horizontal asymptote.0740

Here, the y-axis is an asymptote.0746

All right, let's go ahead and look at the graph of the inverse.0760

And that is going to be very simple, because I know that, since this is the inverse, all I have to do0762

is take the domain and make that the range, and then I take the range and make that the domain.0770

And I am going to check and make sure I have all of these values correctly matched up: 1 and 2...yes, I do.0787

Then, I am going to go ahead and graph them: when x is 0, g(x) is 1 (this is f(x)).0799

When x is 1, g(x) is 2; when x is 3, g(x) is way up here at 8, about here.0805

When x is -1, y is 1/2; -2, 1/4; at -3, it is 1/8.0815

Now, I know that I have an exponential function here that I am graphing.0827

So, as expected, the x-axis is an asymptote for g(x), written in this exponential form.0833

So, looking at what this is saying: the vertical asymptote here is at x = 0, so x will never cross this axis.0855

It will never become negative for f(x), and that makes sense.0869

x can never be negative, because there is no value of y that I can take...2 to some value...and get a negative back.0873

So, x cannot be negative, because there is no possible value of y that would turn 2 into a negative number.0884

Therefore, the domain of f(x) is restricted to values greater than 0 (to positive values).0890

Now, since this is the inverse, and the range of this g(x) is going to be the same as the domain of that,0898

that means that the range of g(x) is going to just be all positive numbers.0904

To sum up properties: the graph of a logarithmic function f(x) shows that f(x) is continuous and one-to-one.0910

There are no gaps; there are no discontinuities; also, you can take that graph that we just did and try the vertical line test.0917

No matter where you drew a vertical line, it will only cross the curve of f(x) once.0924

Therefore, this is a function; there is a one-to-one relationship between the values of x and the values of y.0930

We just discussed why the domain of f(x) must be all positive real numbers.0936

The domain cannot include a negative number, because there would not be any value for y that would give you a negative number.0942

You cannot end up with a negative value for x.0953

The range, however, is all real numbers; so y can be a negative number--I can have -2 here, or something.0956

We also saw that the y-axis is an asymptote, and that the graph is going to approach that axis, but it is never actually going to reach it.0962

And finally, just sketching this back out again, the graph of f(x) looks like this.0973

This is the graph of f(x), and I used log2(x) for that.0990

And the y-intercept was at (1,0), and illustrated here, the y-axis is an asymptote.0999

The domain is only positive numbers; however, the range is all real numbers.1007

Since f(x) equals bx, and g(x) is logb(x), and they are inverses of each other,1017

we can end up with the identity function when we use a composite function, or composition of functions.1023

Recall that, when we talked about composition of functions, we ended up with something like this: f composed with g equals f(g(x).1030

Applying that up here, that is going to give me f(logb(x)) = b...and this is going to be my x;1044

so for x, I am going to insert this: logb(x).1055

And since these are inverses, this two essentially cancel each other out; and I am just going to end up with my x back (the identity function).1058

g composed with f should do the same thing: this is g(f(x)) = g of f(x), which is bx.1070

g here is logb(x), so I am going to take logb, and for x right here, I am going to substitute in bx.1080

And this will work as an identity function, giving me x back.1092

And the inverse property is going to be very helpful to us, as we work on solving equations involving logarithms and exponents.1099

Let's take a look at some methods for solving equations, starting with just very simple ones, and then advancing to more complex equations.1111

Logarithmic equations are defined as those that contain one or more logarithms with a variable in them.1119

The definition can be used to solve simple logarithmic equations.1125

First, recall what the definition of the logarithm is: logb(x) = y if this base, b, when raised to the y power, generates x.1130

This definition can be used to solve very simple logarithmic equations where there is a log only on one side of the equation.1142

By log, I mean a log containing a variable.1149

logb(x + 4) = 3: recall that I said that, if you have a logarithm, you can often use the exponential form to help you solve the equation.1154

And later on, we will see that, if you have the exponential form, you can use the log to help you solve equations.1166

So, thinking this out: I know that I have the base equal to 2 and that x = x + 4 and y = 3.1173

So, I can rewrite this in this form: my base 2, to the third power, equals x, which is x + 4.1180

Now, I have something I can solve: 23 is 8; 8 = x + 4.1190

All I have to do is subtract 4 from both sides, and I get x = 4.1195

Now, you have to be careful when you are working with logs, because we can't take the log of a negative number.1202

So, I am going to just check back in my original here and see...1208

If I put this 4 in here, I am going to get log2(4 + 4), which is log2(8).1213

And that is fine, because that is a positive number.1223

I went ahead and solved this, and then I just double-checked that I ended up with a value that is allowed.1226

Logarithmic inequalities are a similar idea: these are inequalities that involve logarithms.1237

And if b is greater than 1, and x is a positive number, and I have logb(x) > y...1242

(again, we are talking about just a very simple situation, where there is a log on only one side), then x is greater than by.1251

Look at what we did here: we went from the log form to the exponential form.1260

Recall that, with equations, it would look like this: we are doing this same thing, only we are doing it with inequalities.1270

And this relationship still holds up: if the logb(x) > y, then x must be greater than by.1280

It is a little more complicated with less than: so let's just start out talking about greater than, and illustrating it with an example.1294

I am starting out with log3(x + 4) > 2.1300

I know that, if this is true, then I can convert it to this form to solve, because this relationship will hold up.1312

What this is telling me is that, if I take my base equal to 3, and x is equal to (x + 4), and y is equal to 2, I can convert it to this form.1322

So, x is right here: x + 4 is greater than my base b, 3, raised to the second power.1338

Now I can go ahead and solve: x + 4 > 9, so x > 5.1350

I always have to be careful with logs, and make sure that I don't end up taking the log of a negative number.1356

So, log3(x + 4): if I put 5 in here, or slightly more than 5 (x is greater than 5)...say 5.1, I am going to get, say, 9.1, which is positive.1364

So, I am not worried about taking the log of a negative number, because in order to get a negative number,1377

you would have to have something that would be less than 4, and that is not going to occur here.1382

So, that is fine: it is more complicated with less than, and let's look at why.1387

For less than, let's say I had log4(x - 1) < 3.1391

I can't just come up with some number, some value, "x is less than 2," because the problem is:1399

then I could get into very, very small numbers--very negative numbers--1405

where I could end up taking the log of a negative number, which we are not doing.1410

I have to put a restriction on the other side; I have to make sure that x is greater than 0.1416

So, when I convert to this form, instead of just x < by, I need to say, "but it is also greater than 0."1422

So, the base here is 4; I have my 0 here; x in this case is x - 1; the base is 4; and y is 3.1436

So, looking at this: the base is 4; x is x - 1; and y is equal to 3.1453

Therefore, 0 is less than x - 1; 4 times 4 is 16, times 4 is 64.1460

I am going to add a 1 to both sides; and this is going to give me that x is greater than 1, but less than 65.1470

It has put a restriction, a lower limit, on this.1479

OK, we just discussed solving equations with a logarithm on one side.1482

In order to solve equations with logarithms on both sides, we use the following property.1486

If the base is greater than 0, but not equal to 1, then if you have a logarithmic equation1491

with a log on each side, and whose bases are the same, then x must equal y.1497

So, the restriction here is that the bases need to be the same.1505

And you might remember back to working with exponential equations: we said that, if you have an exponential equation,1507

and the bases are the same, then the exponents must be equal.1513

It is similar logic here, which is not surprising, since this is just a notation for working with exponents.1516

For example, log4(3x - 1) = log4(x + 5).1524

Since these bases are the same, then in order for this equation to be valid, 3x - 1 has to equal x + 5.1533

This leaves me with a linear equation that I can easily solve.1544

First, I am going to add a 1 to both sides to get 3x = x + 6; then I am going to subtract an x from each side to get 2x = 6.1547

Then I will divide both sides by 2 to get x = 3.1559

Remember, when working with logs, it is very important to check and make sure that you have a valid solution,1563

because you can't take the log of something that is negative.1567

So, I need to make sure that this expression is not going to end up negative, and this expression is not going to end up negative.1572

So, I am going to plug in this value and see what happens.1579

For this first one, I am going to get log4(9 - 1), which is log4(8); that is valid.1584

Now, this should be the same in here, since these two are equivalent; but we will just double-check it anyway.1592

log4(x + 5) would be log4(3 + 5), or log4(8).1597

And since that is taking a log of a positive number, that is allowed, and this is a valid solution.1604

When working with inequalities with logarithms on both sides, we can solve these inequalities with the following formula.1612

If b is greater than 1, then if you have a log with a certain base, logb(x) > logb(y),1619

then this relationship is true if and only if x is greater than y.1630

The relationship between these two is maintained as long as x is greater than y; so that is a given.1634

And logb(x) is less than logb(y) if and only if x is less than y.1642

This is similar to the logic that we use when solving logarithmic equations with one log on each side.1648

We are doing the same idea, but with inequalities.1655

You need to make sure that you exclude solutions that would require taking the log of a number less than or equal to 0 in the original inequality.1658

So, we are going to look for excluded values at the end, and make sure that we remove those from the solution set.1666

Solutions to the inequality need to actually make the inequality valid.1674

And they need to be not part of the excluded values.1681

I will illustrate that right now: log5(3x + 2) ≥ log5(x - 4).1685

Since the bases are the same, I know that this needs to be greater than or equal to this expression.1698

So, 3x + 2 ≥ x - 4.1706

Therefore, just solving this inequality is going to give me 3x ≥ x - 6; 3x - x ≥ -6; 2x ≥ -6.1716

Divide both sides by 2; I have x ≥ -3.1734

And it might be tempting to stop there; but I need to go back and look at the original.1738

I need to make sure that I am taking only values (in my solution set) that are not excluded.1745

When we were working with equations, it was simple: we would get something like x = 2.1752

And then, we just had to plug it in here and make sure that that was valid; we were fine.1755

Here, we have a whole solution set; so I have to use inequalities to find excluded values,1760

or to find what x must be to end up with a valid solution set.1767

So, I am going to look at log5(3x + 2): for this to be valid, I need for this in here to be greater than 0.1776

All right, that would give me 3x > -2, or x > -2/3.1795

So, this log right here will be valid, as long as x is greater than -2/3.1808

If it is smaller than that, I will end up having an excluded value.1819

Let's look at this log: log5(x - 4): in order for this to be valid, I need to have x - 4 be greater than 0.1825

That means that x would have to be greater than 4.1838

If x is some value less than 4, I will end up taking the log of a negative number, or 0; that is not allowed, so that would not be a valid solution.1840

Now, look at my solution set: my solution set says that x has to be greater than or equal to -3.1849

But that would encompass values that are too small--excluded values--values that are not allowed, like...1855

I could end up with 0, which would then make this -4, and then I would be taking the log of a negative number.1866

So, I have to have a solution set that meets this criteria, this, and this--the most restrictive set.1872

x needs to be greater than or equal to -3, greater than -2/3, and greater than 4.1880

So, I need to go with this: x > 4--that is the solution set.1886

So, when there is an inequality with a logarithm on both sides with the same base,1892

you put this expression on the left side, keep the inequality the same, put this expression on the right, and solve.1898

Then, find excluded values, and make sure that your solution set does not include excluded values.1907

All right, in the first example, we have a logarithmic equation that only has a log on one side.1914

And recall that we can use the definition of logarithms to solve this.1919

logb(x) = y if the base, raised to the y power, equals x.1923

Therefore, I can rewrite this in this form...so, just rewriting it as it is, here the base is equal to 2;1930

x is equal to x3 + 3; and y is equal to 7.1945

So, writing it in this form would give me 27 = x3 + 3.1950

Let's figure out what 2 to the seventh power is: 23 is 8, times 2 is 16, times 2 is 32, times 2 is 64, and then times 2 is going to give me 128.1955

So, this is 2 to the third, fourth, fifth, sixth, seventh: so 27 is going to be 128.1971

All right, I am rewriting this as 128 = x3 + 3.1979

I am going to subtract 3 from both sides: x3 = 125.1986

The cube root of x3 is x, and the cube root of 125 is 5, because 5 times 5 is 25, times 5 is 125.1994

Now, I also have to make sure I don't end up with a negative value inside the log.2004

So, I am going to check this solution to make sure it is valid.2008

x base 2...I am going back to the original...x cubed plus 3...I need to make sure that this is not negative when I use the solution.2011

log2(53 + 3)...well, I know that this is going to be positive; so that is fine--this is a valid solution.2019

All right, log2(2x - 2) < 4--now I am working with an inequality that has a log on only one side.2034

And I am going to go back and think about my definition of logs--that logb(x) = y if by = x.2045

And by rewriting this using the exponential form, I can much more easily solve it.2057

Now, I am working with less than; so I need to recall that, if logb(x) < y, then x is greater than 0, but less than by.2067

If I didn't put this restriction here, I could end up with a value that is too small, and that would be excluded.2090

All right, so rewriting this in this form: I am going to have the 0 here, and x is greater than that.2100

x is 2x - 2...is less than the base, which is 2, raised to 4.2107

I am rewriting this in an exponential form, but making sure that I have this restriction, since we are working with less than.2122

So, 0 is less than 2x - 2; 2 times 2 is 4, times 2 is 8, times 2 is 16.2127

Now, I am going to add 2 to both sides to get 2x is greater than 2, but less than 18;2136

and then divide both sides by 2 to give me x is greater than 1 and less than 9.2143

So, I solved this by using this property of logarithmic inequalities, and making sure that I had the 0, since I am working with less than--2150

that I had the restriction that this expression is greater than 0,2160

so that this in here (what goes inside for the log) does not end up being negative.2165

I don't need to worry about that with greater than.2173

This time, we are going to be solving a logarithmic equation in which there is one log on each side of the equation.2177

Recall that we talked about the property: if there are two logs with the same base, logb(x) = logb(y)--2183

the same bases--then for this equation to be valid, x has to equal y.2193

So, I have log6(x2 - 6) = log6(x).2199

Since these are both base 6, I can just say, "OK, x2 - 6 = x."2206

This is just a quadratic equation: I move the x to the left, and I am going to solve by factoring, just as I would another quadratic equation.2214

This is x + a factor, and then x minus a factor of 6, equals 0.2225

Factors of 6 are 1 and 6, and 2 and 3; and I need them to add up to -1.2232

So, -3 + 2 = -1; I put the 2 here and the 3 here.2238

Using the zero product property, I can solve this, because x + 2 = 0 and x - 3 could equal 0.2247

And either way, this is going to become 0.2255

x = -2 for this; and another solution is x = 3.2259

Now, I have two possible solutions; I need to check these.2263

Let's go up here and look at this, with the -2.2273

I have log6(x2 - 6); if x = -2, then I am going to end up with log6(4 - 6), or log6(-2).2277

That is not valid; I could have also just looked up here and said, "OK, if x equals -2, I would be taking log6(-2)."2292

So, that is not valid; the solution is not valid.2299

Let's try x = 3: well, if x equals 3, and I take the log base 6 of 3, that is OK.2308

Let's check this one out: log6(x2 - 6): log base 6...and we are letting x equal 3 here.2314

of 3 squared minus 6; so that is log6(9 - 6), or log6(3), which is positive.2329

That is allowable; log6(3) is allowable; so these are both allowable.2339

Therefore, the solution is simply x = 3.2344

We came up with two solutions; one was extraneous; we checked and found that we have one valid solution, which is x = 3.2351

Example 4: Solve this inequality that involves a logarithm on each side of the inequality.2360

And I am going to recall that, if the bases are the same (which they are), then logb(x) > logb(y) only if x > y.2367

x must be greater than y; that relationship has to hold up.2381

I am just going to go ahead and look at what I have in here, which is x2 - 9 > x + 3.2385

And I am going to move the 3 to the right by subtracting a 3 from both sides.2394

So, x2 - 9 - 3 > x; so x2 - 12 > x.2401

I am going to subtract an x from both sides; it is going to give me x2 - x - 12 > 0 (the 0 will be left behind).2410

So, this gets into material that we learned earlier on in the course; and it is a little bit conceptually complex.2422

But if you just think it out, you can solve this.2428

Let's factor this out; that is always a good first step.2433

This gives me (x - 4) (x + 3), because the outer terms are -3x - 4x; the inner terms will give me -x; and -4 times 3 is -12.2439

If I wanted to graph this out, I could say, "OK, (x - 4)(x + 3)...let's turn it into the corresponding equation and find the roots."2456

x - 4 = 0, so x = 4; I am finding the roots of this corresponding quadratic equation, just as we did earlier in the course when we were solving quadratic inequalities.2465

Also, x - 3 = 0 would satisfy this equation if x = 3.2477

Actually, that is x + 3 = 0, so x = -3.2487

What this is telling me is that the corresponding quadratic equation has roots at -3 and 4; the zeroes are there.2493

Something else I know is that the leading coefficient here is positive; so this is going to be a parabola that faces upward.2505

And this is really all I need to solve this.2515

And then, I need to go back to my inequality and say, "OK, if this graph looks like this,2518

and what I want is this function, the y-values, to be greater than 0, where are those going to be?"2523

The graph crosses the x-axis right here, at -3: when x is -3, y is 0.2533

For all values of x that are more negative than -3, y is positive.2540

So, for this portion of the graph, when x is less than -3, y is greater than 0, which is what I want.2545

I look over here, and the graph crosses the x-axis at x = 4; so for all values of x that are greater than 4, y is greater than 0.2560

Therefore, in order to satisfy this inequality, x can be less than -3, or x can be greater than 4.2574

Now, I can't stop there and say that I have solve this inequality, because the problem is2587

that I need to make sure that I am not dealing with excluded values.2591

So, let's look at what the excluded values are going to be.2595

If I have log7(x2 - 9), I need for this x2 - 9 to be greater than 0.2597

So, if I factor that, I am going to get (x + 3) (x - 3) > 0.2609

In order for (let's rewrite this a little bit better)...(x + 3) (x - 3) needs to be greater than 0.2620

Therefore, x + 3 needs to be greater than 0, so x needs to be greater than -3.2633

And x - 3 needs to be greater than 0; so x needs to be greater than 3.2639

This is a restriction, and this is a restriction, that if I don't meet these restrictions in my solution set,2647

I am going to end up with excluded values, and it is going to be invalid.2658

So, I am going to look at my solution set: x > 4 meets these criteria, that x > -3 and x > 3.2662

And I don't need to worry about this one, because I already have that factor covered right here: x + 3 > 0.2670

This meets the criteria: it doesn't include any excluded values.2675

However, when I look at x < -3, it doesn't meet all of the criteria, because here, this says x has to be greater than -3.2679

If I take a value, -4, it is not going to meet this criteria; it is not going to meet this criteria, either (that x needs to be greater than 3).2689

Therefore, this is not valid; and my solution is just going to be x > 4,2698

because it solves the inequality--it satisfies the inequality--and it doesn't include excluded values.2705

That one was pretty difficult: first you had to realize that, with the same bases, then you could just take this expression2713

and say that it is greater than this expression and go about solving the inequality.2721

That was a little bit tricky, because then you had to think about what this meant,2726

solve this quadratic inequality, and then realize that x < 3, or values of x that are greater than 4,2730

would satisfy the inequality, but there were some excluded values as part of this solution set.2736

So, we had to just go with x > 4.2742

That concludes this lesson of Educator.com on logarithmic equations and inequalities; and thanks for visiting!2746