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Chapter 13: Problem 26

Write as the sum or difference of logarithms and simplify, if possible. Assume all variables represent positive real numbers. $$\log _{8} 64^{12}$$

Short Answer

Expert verified
The simplified form of the expression \(\log_{8}(64^{12})\) is 24.

Step by step solution

01

1. Using exponent property of logarithms

The exponent property of logarithms states that for any positive number a, b 鈮� 1, and any positive x and n, we have \(\log _{a}(b^{n}) = n\log _{a}(b)\). So putting the given logarithm in this format, we get: \[ \log_{8}(64^{12}) = 12\log_{8}(64) \]
02

2. Converting base 8 to base 2

Next, we'll convert the base-8 logarithm to a base-2 logarithm using the change of base formula: given \(\log_b(a) = c\), we can write it as \(\log_b(b^c) = c\). Now we write 8 as 2 to the power of 3, so: \[ 12\log_{8}(64) = 12\log_{2^3}(64) \]
03

3. Using change of base formula

Now, we'll use the change of base formula on the base-2 logarithm: \(\log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)}\). Here, we will use base 2, a=2^3, and b=64: \[ 12\log_{2^3}(64)= 12\frac{\log_{2}(64)}{\log_{2}(2^3)} \]
04

4. Simplifying the expression

Now, we can further simplify the terms in the numerator and denominator: \[ 12\frac{\log_{2}(64)}{\log_{2}(2^3)} = 12\frac{\log_{2}(2^6)}{\log_{2}(2^3)} \] Since \(\log_2({2^n}) = n\), we can simplify this expression to: \[ 12\frac{6}{3} \]
05

5. Final simplification

Now we can proceed with the final simplification: \[ 12\frac{6}{3} = 12\cdot 2 = 24 \] So, the simplified form of the expression \(\log_{8}(64^{12})\) is 24.

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