Question

what is the approximate value of x in the equation below.
\log_{\frac{3}{4}} 25 = 3x - 1
\bigcirc -3.396
\bigcirc -0.708
\bigcirc 0.304
\bigcirc 0.955

Explanation:

Step1: Recall logarithm change of base formula

The change of base formula for a logarithm $\log_b a$ is $\frac{\ln a}{\ln b}$ (or $\frac{\log_c a}{\log_c b}$ for any positive $c
eq1$). So for $\log_{\frac{3}{4}} 25$, we can write it as $\frac{\ln 25}{\ln \frac{3}{4}}$.

Step2: Calculate the logarithm value

First, calculate $\ln 25\approx3.2189$ and $\ln \frac{3}{4}=\ln 3 - \ln 4\approx1.0986 - 1.3863=-0.2877$. Then $\log_{\frac{3}{4}} 25=\frac{3.2189}{-0.2877}\approx - 11.188$. Wait, no, wait, the equation is $\log_{\frac{3}{4}} 25 = 3x - 1$. Wait, maybe I made a mistake in the base. Wait, the base is $\frac{3}{4}$, which is less than 1, so the logarithm of a number greater than 1 (25) should be negative, which matches. Now, solve for $x$:
Start with $\log_{\frac{3}{4}} 25 = 3x - 1$. Let's compute $\log_{\frac{3}{4}} 25$ using change of base formula: $\log_{\frac{3}{4}} 25=\frac{\log 25}{\log \frac{3}{4}}=\frac{\log 5^2}{\log 3 - \log 4}=\frac{2\log 5}{\log 3 - \log 4}$. We know that $\log 5\approx0.6990$, $\log 3\approx0.4771$, $\log 4\approx0.6021$. So numerator: $2\times0.6990 = 1.398$. Denominator: $0.4771 - 0.6021=-0.125$. Then $\log_{\frac{3}{4}} 25=\frac{1.398}{-0.125}\approx - 11.184$? Wait, that can't be right. Wait, no, wait the base is $\frac{3}{4}=0.75$, so let's use natural logarithm correctly. $\ln 25 = \ln 5^2 = 2\ln 5\approx2\times1.6094 = 3.2188$. $\ln 0.75=\ln \frac{3}{4}\approx - 0.2877$. So $\log_{0.75}25=\frac{3.2188}{-0.2877}\approx - 11.188$. Then the equation is $-11.188 = 3x - 1$. Wait, that would give $3x=-11.188 + 1=-10.188$, $x\approx - 3.396$. Wait, but that's one of the options. Wait, maybe I messed up the equation? Wait, the original equation is $\log_{\frac{3}{4}} 25 = 3x - 1$? Let me check the options. One of the options is -3.396. Let's verify:

If $x = - 3.396$, then $3x - 1=3\times(-3.396)-1=-10.188 - 1=-11.188$. And $\log_{\frac{3}{4}}25=\frac{\ln 25}{\ln \frac{3}{4}}=\frac{3.2189}{-0.2877}\approx - 11.188$. So that's correct.

Wait, but let's do it step by step again.

1. Start with the equation: $\log_{\frac{3}{4}} 25 = 3x - 1$
2. Solve for $x$: First, add 1 to both sides: $\log_{\frac{3}{4}} 25 + 1 = 3x$
3. Then divide both sides by 3: $x=\frac{\log_{\frac{3}{4}} 25 + 1}{3}$
4. Calculate $\log_{\frac{3}{4}} 25$: Using change of base formula, $\log_{\frac{3}{4}} 25=\frac{\ln 25}{\ln \frac{3}{4}}\approx\frac{3.2189}{-0.2877}\approx - 11.188$
5. Then $\log_{\frac{3}{4}} 25 + 1\approx - 11.188 + 1=-10.188$
6. Then $x=\frac{-10.188}{3}\approx - 3.396$

Answer:

-3.396 (corresponding to the first option: -3.396)