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 nonpositive value in the original equation or inequality. Exclude such values from the solution set.
Logarithms and Logarithmic Functions
 Recall that log_{b}x = y is the same as b^{y} = x
 b = 4
 y = 3
 x = x^{4} − 17
 Rewrite into exponential form.
 4^{3} = x^{4} − 17
 64 = x^{4} − 17
 Solve
 x^{4} = 81
 Recall that log_{b}x = y is the same as b^{y} = x
 b = 4
 y = 3
 x = x^{4} − 192
 Rewrite into exponential form.
 4^{3} = x^{4} − 192
 64 = x^{4} − 192
 Solve
 x^{4} = 256
 Recall that log_{b}x = y is the same as b^{y} = x
 b = 4
 y = 4
 x = x^{4} − 369
 Rewrite into exponential form.
 4^{4} = x^{4} − 369
 256 = x^{4} − 369
 Solve
 x^{4} = 625
 Recall that log_{b}x = y is the same as b^{y} = x
 b = 2
 y = 4
 x = x^{3} − 713
 Rewrite into exponential form.
 2^{4} = x^{3} − 713
 16 = x^{3} − 713
 Solve
 x^{3} = 729
 Notice that since the bases are the same, you can use the following property
 log_{b}x = log_{b}y; then x = y
 log_{5}(x^{2} − 12) = log_{5}(x)
 x^{2} − 12 = x
 x^{2} − x − 12 = 0
 Factor
 x^{2} − x − 12 = (x − )(x + )
 x^{2} − 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 log_{5}(x^{2} − 12) = log_{5}(x) log_{5}(x^{2} − 12) = log_{5}(x) log_{5}(4^{2} − 12) = log_{5}(4) log_{5}(( − 3)^{2} − 12) = log_{5}( − 3) log_{5}(4) = log_{5}(4) log_{5}( − 3) = log_{5}( − 3) x = 4 is valid Not valid, you cannot have negative logarithms.
 Notice that since the bases are the same, you can use the following property
 log_{b}x = log_{b}y; then x = y
 log_{5}(x^{2} + 4) = log_{5}( − 5x)
 x^{2} + 4 = − 5x
 x^{2} + 5x + 4 = 0
 Factor
 x^{2} + 5x + 4 = (x + )(x + )
 x^{2} + 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 log_{5}(x^{2} + 4) = log_{5}( − 5x) log_{5}(x^{2} + 4) = log_{5}( − 5x) log_{5}(x^{2} + 4) = log_{5}( − 5x) log_{5}(( − 1)^{2} + 4) = log_{5}( − 5( − 1)) log_{5}(( − 4)^{2} + 4) = log_{5}( − 5( − 4)) log_{5}(5) = log_{5}(5) log_{5}(20) = log_{5}(20) x=1 is valid x=4 is valid
 Recall that log_{b}x = y can be written in exponential form as b^{y} = x.
 Don't forget that there's always a restriction when working with logs, namely
 log_{b}x < y; then 0 < x < b^{y}
 Solve
 0 < x < b^{y}
 0 < 5x − 3 < 3^{3}
 0 < 5x − 3 < 27
 3 < 5x < 30
 Recall that log_{b}x = y can be written in exponential form as b^{y} = x.
 Don't forget that there's always a restriction when working with logs, namely
 log_{b}x < y; then 0 < x < b^{y}
 Solve
 0 < x < b^{y}
 0 < 9x − 18 < 12^{2}
 0 < 9x − 18 < 144
 18 < 9x < 162
 Recall that log_{b}x = y can be written in exponential form as b^{y} = x.
 Don't forget that there's always a restriction when working with logs, namely
 log_{b}x < y; then 0 < x < b^{y}
 Solve
 0 < x < b^{y}
 0 < 10x + 9 < 7^{2}
 0 < 10x + 9 < 49
 − 9 < 10x < 40
 Since the base of the exponents is the same, we can take what is inside the parenthesis outside
 8 + 3x < x^{2} − 2
 Move everything to one side of the inequality
 0 < x^{2} − 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
 log_{3}(8 + 3x) < log_{3}(x^{2} − 2)
 8 + 3x > 0 3x > − 8 x > − [8/3]
 x^{2} − 2 > 0 x^{2} > 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
*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.
Answer
Logarithms and Logarithmic Functions
Lecture Slides are screencaptured 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
 What are Logarithms?
 Logarithmic Functions
 Graph of the Logarithmic Function
 Properties
 Inverse Property
 Equations
 Inequalities
 Equations with Logarithms on Both Sides
 Inequalities with Logarithms on Both Sides
 Example 1: Solve Log Equation
 Example 2: Solve Log Equation
 Example 3: Solve Log Equation
 Example 4: Solve Log Inequality
 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
 YAxis 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
 Example 4: Solve Log Inequality 39:19
Algebra 2
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: log_{b}(x).0021
Let's start out with log_{b}(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 log_{b}(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
is: exponential expressions can help you solve logarithmic equations, and logarithmic expressions can help you solve exponential equations.0098
Given log_{4}(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 truethat 4^{2} 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 2^{3} = 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 log_{2}(8) = 3.0164
We can also use this for slightly more complicated situations, such as log_{5}(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 1/5^{3}, 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 log_{3}(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 3^{4} 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) = log_{b}(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) = log_{b}(x)0347
is the inverse of the exponential equation, expression, or function g(x) = b(x).0355
So, f(x) = log_{b}(x) is the inverse of g(x) = b^{x}.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 log_{2}(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 log_{2}(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, 2^{0} = 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; 2^{2} = 4; therefore y = 2.0451
Let's let x be 8: 2 to some power y equals 8: well, 2^{3} 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 2^{2}, is 4.0509
When x is 3, y is 2^{3}: 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) = log_{2}(x), and for g(x) = 2^{x}, the inverse.0572
So, let's keep going with those, but add on some values.0584
Recall that I said, if f(x) = log_{2}(x), then if I take 2^{y}, 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 2^{y} = 1.0610
And I said that y must be 0; when x is 2, 2^{y} = 2, so y must be 1.0617
And I went on and did a couple of other values, and I did 8: I said 2^{y} = 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 2^{y} = 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: 2^{y} = 1/4if I took 2^{2}, I would get 1/4, so y is 2.0668
1/8: using that same logic, 2^{3} 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 yaxis is a vertical asymptote.0730
And we talked earlier about the graphs of exponential functions: we saw that the xaxis formed a horizontal asymptote.0740
Here, the yaxis 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 xaxis 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 onetoone.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 onetoone 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 numberI can have 2 here, or something.0956
We also saw that the yaxis 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 log_{2}(x) for that.0990
And the yintercept was at (1,0), and illustrated here, the yaxis is an asymptote.0999
The domain is only positive numbers; however, the range is all real numbers.1007
Since f(x) equals b^{x}, and g(x) is log_{b}(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(log_{b}(x)) = b...and this is going to be my x;1044
so for x, I am going to insert this: log_{b}(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 b^{x}.1070
g here is log_{b}(x), so I am going to take log_{b}, and for x right here, I am going to substitute in b^{x}.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: log_{b}(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
log_{b}(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: 2^{3} 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 log_{2}(4 + 4), which is log_{2}(8).1213
And that is fine, because that is a positive number.1223
I went ahead and solved this, and then I just doublechecked 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 log_{b}(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 b^{y}.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 log_{b}(x) > y, then x must be greater than b^{y}.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 log_{3}(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, log_{3}(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 log_{4}(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 numbersvery negative numbers1405
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 < b^{y}, 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, log_{4}(3x  1) = log_{4}(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 log_{4}(9  1), which is log_{4}(8); that is valid.1584
Now, this should be the same in here, since these two are equivalent; but we will just doublecheck it anyway.1592
log_{4}(x + 5) would be log_{4}(3 + 5), or log_{4}(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, log_{b}(x) > log_{b}(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 log_{b}(x) is less than log_{b}(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: log_{5}(3x + 2) ≥ log_{5}(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 log_{5}(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: log_{5}(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 smallexcluded valuesvalues 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 thisthe 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 > 4that 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
log_{b}(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 x^{3} + 3; and y is equal to 7.1945
So, writing it in this form would give me 2^{7} = x^{3} + 3.1950
Let's figure out what 2 to the seventh power is: 2^{3} 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 2^{7} is going to be 128.1971
All right, I am rewriting this as 128 = x^{3} + 3.1979
I am going to subtract 3 from both sides: x^{3} = 125.1986
The cube root of x^{3} 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
log_{2}(5^{3} + 3)...well, I know that this is going to be positive; so that is finethis is a valid solution.2019
All right, log_{2}(2x  2) < 4now 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 logsthat log_{b}(x) = y if b^{y} = 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 log_{b}(x) < y, then x is greater than 0, but less than b^{y}.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 than2150
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, log_{b}(x) = log_{b}(y)2183
the same basesthen for this equation to be valid, x has to equal y.2193
So, I have log_{6}(x^{2}  6) = log_{6}(x).2199
Since these are both base 6, I can just say, "OK, x^{2}  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 log_{6}(x^{2}  6); if x = 2, then I am going to end up with log_{6}(4  6), or log_{6}(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 log_{6}(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: log_{6}(x^{2}  6): log base 6...and we are letting x equal 3 here.2314
of 3 squared minus 6; so that is log_{6}(9  6), or log_{6}(3), which is positive.2329
That is allowable; log_{6}(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 log_{b}(x) > log_{b}(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 x^{2}  9 > x + 3.2385
And I am going to move the 3 to the right by subtracting a 3 from both sides.2394
So, x^{2}  9  3 > x; so x^{2}  12 > x.2401
I am going to subtract an x from both sides; it is going to give me x^{2}  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 yvalues, to be greater than 0, where are those going to be?"2523
The graph crosses the xaxis 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 xaxis 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 log_{7}(x^{2}  9), I need for this x^{2}  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 inequalityit satisfies the inequalityand 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
1 answer
Last reply by: Dr Carleen Eaton
Sat Nov 7, 2015 6:06 PM
Post by Peter Ke on October 21, 2015
Hi, what is a asymptote? I search it on google and I don't really get it. Can you explain it in an easier way?
0 answers
Post by julius mogyorossy on June 22, 2013
The inverse property for logs is like, 2, times 4, divided by 4, = 2.
0 answers
Post by julius mogyorossy on June 10, 2013
Dr. Eaton, again I don't know why you say x'2x12 is y, when you set the right side of the equation to 0, what I call y, it seems you are making it = x. I think the quadratic equation should not be called y or x, but the determinator, something like that. It just confused me, when you set y to 0, then you said x'2x12 is y. I am not sure why you talk about y at all when showing ex4, it does not seem that y has anything to do with that problem, or why you show a quadratic equation graph line, it seems you should have only showed a x dimension line graph. That problem is about finding values for x that make the inequality true, nothing to do with y. I kept thinking there was something I was not seeing. Again the quadratic equation factored said that part of the solution set was x>3, not, x<3, as you said, what you would call an invalid solution, I guess, and it said x>4 is possibly part of the solution set, which is true. Yes, you talked about y, but it really seems it is just about x and x, I know you said one of the x's is a y, but really for this problem it just seems like it is about x and x, it seems that is the better way to see it so as not confuse it with a regular quadraticinequality problem. It really seems the y values for this problem are irrelevant. The equation is set to 0 not to find the roots, the y's=0, but to find the solutions that also do not violate the rule to not take the root of 0 or a #<0, isn't that the case. For on kind of problem you set it to 0 to find the roots, for another to make sure not to take the log of 0 or a #<0. It seems you unnecessarily factored x'29 and x+3 twice, the first time when you combined them into a quadratic equation, killing two birds with one stone, so to speak, then again separately, unnecessarily it seemed, is this true.
1 answer
Last reply by: Dr Carleen Eaton
Sun Mar 11, 2012 7:16 PM
Post by Jeff Mitchell on March 7, 2012
Dr Carleen,
In example IV, you where checking for exclusions and said
(x+3)(x3)>0
X+3>0
X>3
X3>0
X>3
and you indicated 3 is not a valid solution
but wouldn't it be true that if x=4;
(4+3)(43)=(1)(7)>0 which is valid?
1 answer
Last reply by: Dr Carleen Eaton
Mon Nov 7, 2011 8:49 PM
Post by Jonathan Taylor on November 6, 2011
Dr Carleen would u explain how either u subtract or add x+4>9(equalities)
x=5
1 answer
Last reply by: Dr Carleen Eaton
Fri Jun 10, 2011 1:01 AM
Post by Manuel Gonzalez on June 7, 2011
aren't roots the opposite of exponents?
0 answers
Post by Edgar Rariton on March 6, 2011
The definition of a logarithm in this video begins with "First the restrictions...". That's one way to do it, but perhaps explaining that logarithms are the "opposite" of exponentials would have been better.
0 answers
Post by Wade Sias on January 27, 2011
There needs to be more complex problems like
lnx + lnx(x3) <= ln4
I feel the examples are to easy and I am struggling with these longer problems.
1 answer
Last reply by: Suhani Pant
Sat Jul 27, 2013 12:16 PM
Post by Dr Carleen Eaton on August 20, 2010
Correction to Example III at 38:42
When checking the potential solution x = 3, I should have written the original equation as log base6(x squared 6) NOT
log base6 (x squared  3). With the correct equation, substituting x = 3 would give:
logbase6(3 squared  3)
logbase6 (93)
logbase6(3)
Therefore, x = 3 is a valid solution.
I apologize for any confusion my error may have caused.