# Reading: Logarithmic Properties

In the previous section, we derived two important properties of logarithms, which allowed us to solve some basic exponential and logarithmic equations.

David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, " Chapter 4: Exponential and Logarithmic Functions," licensed under a CC BY-SA 3.0 license.

## Properties of Logs

**Inverse Properties****:****Exponential Property****:****Change of Base****:**While these properties allow us to solve a large number of problems, they are not sufficient to solve all problems involving exponential and logarithmic equations.## Properties of Logs

**Sum of Logs Property****:****Difference of Logs Property****:**It's just as important to know what properties logarithms do*not*satisfy as to memorize the valid properties listed above. In particular, the logarithm is not a linear function, which means that it does not distribute: log(*A*+*B*) ≠ log(*A*) + log(*B*). To help in this process we offer a proof to help solidify our new rules and show how they follow from properties you've already seen. Let*a*= log_{b}(*A*) and*c*= log_{b}(*C*), so by definition of the logarithm,*b*^{a}=*A*and*b*C Using these expressions,^{c}=*AC*=*b*Using exponent rules on the right,^{a}b^{c}*AC*=*b*Taking the log of both sides, and utilizing the inverse property of logs, log^{a+c}*(*_{b}*AC*) log*(*_{b}*b*^{a+c}) =*a*+*c*Replacing*a*and*c*with their definition establishes the result log*(*_{b}*AC*) = log_{b}*A*+ log_{b}*C*The proof for the difference property is very similar. With these properties, we can rewrite expressions involving multiple logs as a single log, or break an expression involving a single log into expressions involving multiple logs.### Example 1

Write log_{3}(5) + log_{3}(8) – log_{3}(2) as a single logarithm. Using the sum of logs property on the first two terms, log_{3}(5) + log_{3}(8) = log_{3}(5 × 8) = log_{3}(40) This reduces our original expression to log_{3}(40) – log_{3}(2) Then using the difference of logs property,### Example 2

Evaluate 2log(5) + log(4) without a calculator by first rewriting as a single logarithm. On the first term, we can use the exponent property of logs to write 2log(5) = log(5^{2}) = log(25) With the expression reduced to a sum of two logs, log(25) + log(4), we can utilize the sum of logs property log(25) + log(4) = log(4 × 25) = log(100) Since 100 = 10^{2}, we can evaluate this log without a calculator: log(100) = log(10^{2}) = 2### Try it Now 1

Without a calculator evaluate by first rewriting as a single logarithm: log_{2}(8) + log_{2}(4)### Example 3

Rewrite as a sum or difference of logs First, noticing we have a quotient of two expressions, we can utilize the difference property of logs to write Then seeing the product in the first term, we use the sum property ln(*x*^{4}*y*) – ln(7) = ln(*x*^{4}) + ln(*y*) – ln(7) Finally, we could use the exponent property on the first term ln(*x*^{4}) + ln(*y*) – ln(7) = 4ln(*x*) + ln(*y*) – ln(7)# Try it Now Answers

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David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, " Chapter 4: Exponential and Logarithmic Functions," licensed under a CC BY-SA 3.0 license.