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Applying Properties of Logarithms

Applying Properties of Logarithms

Author: Colleen Atakpu
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Applying Properties of Logarithms

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Today we're going to talk about applying the properties of logarithms. So we're going to do some examples simplifying logarithmic expressions using different properties of logarithms. So our first example, let's see how we can evaluate a logarithmic expression using the relationship between exponential equations and logarithmic equations. So I have here log base 5 of 1 plus log base 5 of 125 minus log base 5 of 25.

So using my relationship between logarithmic and exponentials, I can rewrite this, the value of this expression. If I say that that's equal to some value x, then I will have 5 to that value x is equal to 1.

Similarly, with the second expression, I can write this as an exponential equation if I say that this expression is equal to x, then I can write an exponential form 5 to the x is equal to 125. And finally here, I can write this as 5 to some value x which is equal to this expression is equal to 25. So now that I have equations for each expression, I can solve each equation for x to evaluate my expression.

So 5 to the x equals 1 means that x has to be equal to 0. The only exponent that will evaluate to 1 is 0. So here x is equal to 0. 5 to some exponent equals 125 means that exponent should be 3, because 5 times 5 times 5 gives me 125. So x equals 3. And finally, 5 to some exponent equals 25 means the exponent should be 2, because 5 squared or 5 to the second power is 25.

So now that I have my values for x, I can simply combine them together from my original expression. So I have 0 plus 3 minus 2. 0 plus 3 is 3 minus 2 is just going to give me 1.

So now let's do an example using the product and quotient properties of logarithms. I've got log of x times y divided by z. I'm going to use the product and quotient properties of logarithms to simplify this. And I know that log of x is equal to 2, log of y is equal to negative 1, and log of z is equal to 4. So knowing those values, I'm first going to rewrite this, as I said, using my product and quotient properties. So I know that I can separate the arguments for the logarithm out using addition and subtraction.

So that would become log of x plus log of y minus log of z. And now substituting the values for these logarithms, log of x is 2, log of y is negative 1, and log of z is 4. So we have 2 plus negative 1 minus 4. 2 plus negative 1 is going to give me a positive 1. And then positive 1 minus 4 is going to give me a negative 3.

So let's do one final example applying the different properties of logarithms. If you're feeling pretty confident, go ahead and pause, try it on your own, and the check back with us later. So we've got the example log base 4 of 4 plus log base 4 of 16. So I'm going to start by applying the product property of logarithms. Since my bases are the same, I can write this as a single logarithm, log base 4 and my argument will be 4 times 16. And this-- 4 times 16-- is equal to 64. So this becomes log base 4 of 64. Now I know that 64 is the same as 8 squared. So I can write this as log base 4 of 8 squared.

Using my power property of logarithms, I know that I can rewrite this. I'll take my exponent and use it as a factor. It'll be multiplying in front of the logarithm. So this is going to become 2 times log base 4 of 8. Now I'm going to use my change of base formula, because I can't evaluate a log with a base 4. I can't evaluate this logarithm in my head.

So using my change of base formula, I'll still have the 2 in front that's multiplying. And I'll have log of 8 over log of 4. Now logarithm without a base written is a common logarithm, which most calculators have that logarithm button on there. And so you can simply type in log of 8 divided by log of 4, and that would give you 1.5, a the value of 1.5. Then we bring down our 2 in front. And then finally, 2 times 1.5 is equal to 3.

So let's go over our key points from today. Here are our properties of logarithms. The product property says that log base b of some product x and y is equal to log base b of x plus log base b of y. The quotient property says that log base b of some quotient x and y is equal to log base b of x minus log base b of y. The power property says that log base b of x to the nth power is equal to n times log base b of x. And the change of base property of logs says that log base b of x is equal to log base a of x divided by log base a of b.

So I hope that these key points and examples helped you understand a little bit more about applying the properties of logarithms. Keep using your notes and keep on practicing and soon you'll be a pro. Thanks for watching.

Notes on "Applying Properties of Logarithms"

Key Formulas

Product Property of Logs: log subscript b left parenthesis x y right parenthesis equals log subscript b x plus log subscript b y

Quotient Property of Logs: log subscript b left parenthesis x over y right parenthesis equals log subscript b x minus log subscript b y

Power Property of Logs: log subscript b x to the power of n equals n times log subscript b x

Change of Base Property of Logs: log subscript b x equals fraction numerator log subscript a x over denominator log subscript a b end fraction

Key Terms

None