Question

question
evaluate:
$log_{64} \frac{1}{2}$
answer
attempt 1 out of c

Explanation:

Step1: Recall logarithm power rule

The power rule of logarithms states that $\log_a b^n = n\log_a b$. We can rewrite the argument of the logarithm as a power of a base related to the logarithm's base. First, note that $64 = 8^2$, so we can rewrite the logarithm as $\log_{64} \frac{1}{2}$. Wait, no, the original is $\log_{64} \frac{1}{2}$? Wait, no, the user's image shows "log base 64 of 1/2"? Wait, no, the text in the image: "Evaluate: $\log_{64} \frac{1}{2}$"? Wait, no, looking at the image, the logarithm is $\log_{64} \frac{1}{2}$? Wait, no, the user's image: the logarithm is $\log_{64} \frac{1}{2}$? Wait, no, let's re-express. Wait, maybe it's $\log_{64} \frac{1}{2}$? Wait, no, let's use the change of base formula or express $\frac{1}{2}$ as a power of 64. Let's let $y = \log_{64} \frac{1}{2}$. By the definition of logarithms, this means $64^y = \frac{1}{2}$. Now, express 64 and $\frac{1}{2}$ as powers of 2. We know that $64 = 2^6$ and $\frac{1}{2} = 2^{-1}$. So substitute these into the equation: $(2^6)^y = 2^{-1}$. Using the power of a power rule, $(a^m)^n = a^{mn}$, so this becomes $2^{6y} = 2^{-1}$. Since the bases are the same and the exponential function is one-to-one, we can set the exponents equal to each other: $6y = -1$. Then solve for $y$: $y = -\frac{1}{6}$. Wait, but maybe I misread the logarithm. Wait, the user's image: is it $\log_{64} \frac{1}{2}$ or $\log_{64} \frac{1}{2}$? Wait, maybe the original is $\log_{64} \frac{1}{2}$? Wait, no, let's check again. Wait, the image shows "log base 64 of 1/2". Wait, but let's confirm. Alternatively, maybe it's $\log_{64} \frac{1}{2}$? Wait, no, let's do the calculation again. Let $y = \log_{64} \frac{1}{2}$. Then $64^y = \frac{1}{2}$. $64 = 2^6$, so $(2^6)^y = 2^{-1}$. Then $2^{6y} = 2^{-1}$. So $6y = -1$ => $y = -\frac{1}{6}$. Wait, but maybe the logarithm is $\log_{64} \frac{1}{2}$? Wait, no, maybe I made a mistake. Wait, let's check with another method. The change of base formula: $\log_a b = \frac{\ln b}{\ln a}$. So $\log_{64} \frac{1}{2} = \frac{\ln \frac{1}{2}}{\ln 64} = \frac{-\ln 2}{\ln 2^6} = \frac{-\ln 2}{6\ln 2} = -\frac{1}{6}$. Yes, that's correct. So the value is $-\frac{1}{6}$. Wait, but maybe the original problem was $\log_{64} \frac{1}{2}$? Wait, the user's image: let's see, the text is "Evaluate: $\log_{64} \frac{1}{2}$"? Wait, no, the image shows "log base 64 of 1/2". So the calculation is as above.

Step1: Let \( y = \log_{64} \frac{1}{2} \)

By definition of logarithm, \( 64^y = \frac{1}{2} \)

Step2: Express in terms of base 2

Since \( 64 = 2^6 \) and \( \frac{1}{2} = 2^{-1} \), substitute:
\( (2^6)^y = 2^{-1} \)

Step3: Apply power of a power rule

\( 2^{6y} = 2^{-1} \)

Step4: Equate exponents (one - to - one property)

\( 6y = -1 \)

Step5: Solve for \( y \)

\( y = -\frac{1}{6} \)

Answer:

\(-\frac{1}{6}\)