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Study Guides > Finite Math

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.

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 = logb(A) and c = logb(C), so by definition of the logarithm, ba = A and bc = C Using these expressions, AC = babc Using exponent rules on the right, AC = ba+c Taking the log of both sides, and utilizing the inverse property of logs, log b(AC) logb(ba+c) = a + c Replacing a and c with their definition establishes the result logb(AC) = logbA + logbC 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) + log3(8) – log3(2) as a single logarithm. Using the sum of logs property on the first two terms, log 3(5) + log3(8) = log3(5 × 8) = log3(40) This reduces our original expression to log 3(40) – log3(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(102) = 2

Try it Now 1

Without a calculator evaluate by first rewriting as a single logarithm: log 2(8) + log2(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( x4y) – ln(7) = ln(x4) + ln(y) – ln(7) Finally, we could use the exponent property on the first term ln( x4) + ln(y) – ln(7) = 4ln(x) + ln(y) – ln(7)

Try it Now Answers

  1. 5

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.