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### 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

_{4}(x

^{4}− 17) = 3

- 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

_{4}(x

^{4}− 192) = 3

- 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

_{4}(x

^{4}− 369) = 4

- 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

_{2}(x

^{3}− 713) = 4

- 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

_{5}(x

^{2}− 12) = log

_{5}(x)

- 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.

_{5}(x

^{2}+ 4) = log

_{5}( − 5x)

- 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

_{3}(5x − 3) < 3

- 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

_{12}(9x − 18) < 2

- 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

_{7}(9x − 18) < 2

- 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

_{3}(8 + 3x) < log

_{3}(x

^{2}− 2)

- 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 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
- 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
- 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
- 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 two*0094

*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 true--that 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} = 1/4--if 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 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 do*0762

*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 log _{2}(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 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 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 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 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 < 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 equation*1491

*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 double-check 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 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

*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 fine--this is a valid solution.*2019

*All right, log _{2}(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 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 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, log _{b}(x) = log_{b}(y)--*2183

*the same bases--then 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 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 is*2587

*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 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 expression*2713

*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

0 answers

Post by John Stedge on June 15 at 03:45:19 PM

at 12:42 you spelled asymptote wrong

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'2-x-12 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'2-x-12 is y. I am not sure why you talk about y at all when showing ex-4, 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 quadratic-inequality 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'2-9 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)(x-3)>0

X+3>0

X>-3

X-3>0

X>3

and you indicated -3 is not a valid solution

but wouldn't it be true that if x=-4;

(-4+3)(-4-3)=(-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(x-3) <= 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 (9-3)

logbase6(3)

Therefore, x = 3 is a valid solution.

I apologize for any confusion my error may have caused.