Question

question
evaluate:
\log_{128} \frac{1}{4}
answer attempt 1 out of 2

Explanation:

Step1: Recall the change of base formula

The change of base formula for logarithms is $\log_{a}b=\frac{\ln b}{\ln a}$ (or $\frac{\log_{c}b}{\log_{c}a}$ for any positive $c
eq1$). Also, we know that $128 = 2^7$ and $4=2^2$. Let's rewrite the logarithm using base 2.
We know that $\log_{a^m}b^n=\frac{n}{m}\log_{a}b$. First, express $128$ and $4$ as powers of 2: $128 = 2^7$ and $4 = 2^2$. So $\log_{128}\frac{1}{4}=\log_{2^7}2^{-2}$.

Step2: Apply the logarithm power rule

Using the rule $\log_{a^m}b^n=\frac{n}{m}\log_{a}b$, here $a = 2$, $m = 7$, $b = 2$, $n=- 2$. So $\log_{2^7}2^{-2}=\frac{-2}{7}\log_{2}2$. Since $\log_{a}a = 1$, $\log_{2}2 = 1$. Then $\frac{-2}{7}\times1=-\frac{2}{7}$? Wait, no, wait. Wait, the original expression is $\log_{128}\frac{1}{4}$, let's do it again. Wait, $\frac{1}{4}=4^{-1}= (2^2)^{-1}=2^{-2}$, and $128 = 2^7$. So $\log_{2^7}2^{-2}$. The formula is $\log_{a^m}b^n=\frac{n}{m}\log_{a}b$. So here, $a = 2$, $m = 7$, $b = 2$, $n=-2$. So $\log_{2^7}2^{-2}=\frac{-2}{7}\log_{2}2=\frac{-2}{7}\times1=-\frac{2}{7}$? Wait, no, wait the original problem is $\log_{128}\frac{1}{4}$, let's use change of base formula: $\log_{128}\frac{1}{4}=\frac{\ln\frac{1}{4}}{\ln128}=\frac{\ln4^{-1}}{\ln2^7}=\frac{- \ln4}{7\ln2}=\frac{- \ln2^2}{7\ln2}=\frac{-2\ln2}{7\ln2}=-\frac{2}{7}$. Wait, but wait, the problem is $\log_{128}\frac{1}{4}$? Wait, the user wrote "log₁₂₈ ¼", let's confirm. Wait, 128 is $2^7$, ¼ is $2^{-2}$. So $\log_{2^7}2^{-2}=\frac{-2}{7}\log_{2}2=\frac{-2}{7}$. But wait, let's check with change of base formula: $\log_{128}\frac{1}{4}=\frac{\log_{2}\frac{1}{4}}{\log_{2}128}=\frac{\log_{2}2^{-2}}{\log_{2}2^7}=\frac{-2\log_{2}2}{7\log_{2}2}=\frac{-2}{7}$. Yes, that's correct. Wait, but the original problem in the image: wait, the user's image shows "log₁₂₈ ¼"? Wait, no, wait the image: "Evaluate: log₁₂₈ ¼" (wait, no, the image has "log₁₂₈ ¼"? Wait, the user's image: let's re - examine. The image says "Evaluate: log₁₂₈ ¼"? Wait, no, the image: "Evaluate: log₁₂₈ ¼" (wait, the user's image: the text is "Evaluate: log₁₂₈ ¼"? Wait, no, looking at the image: "Evaluate: log₁₂₈ ¼" (wait, the vertical text: "log₁₂₈ ¼" – wait, no, the image is rotated? Wait, no, the user's image: the text is "Evaluate: log₁₂₈ ¼" (wait, maybe I misread. Wait, the original problem: "Evaluate: log₁₂₈ (1/4)". Let's do it step by step.

Alternative approach: Let $y=\log_{128}\frac{1}{4}$. By definition of logarithm, $128^y=\frac{1}{4}$. Express 128 and 4 as powers of 2: $128 = 2^7$, $\frac{1}{4}=2^{-2}$. So $(2^7)^y=2^{-2}$. Using exponent rule $(a^m)^n=a^{mn}$, we get $2^{7y}=2^{-2}$. Since the bases are equal, exponents must be equal: $7y=-2$. Solve for $y$: $y =-\frac{2}{7}$.

Answer:

$-\frac{2}{7}$