<s>[INST] How to find $x$ of this equation I am a beginner to logarithms.
By using logarithms, in this equation, I want to find the value of $x$:
$$ \ 3^x = 4^{x-1} \ $$
I am using 10 as base of logarithm.
And I want $x$ in terms of $\log2$ and $\log3$, as far as possible.
 [/INST] As mentioned logarithm use is desired. Here, for a solution in terms of natural logarithms,
\begin{align}
\ln(3^{x}) &= \ln(4^{x-1}) \\
x \, \ln(3) &= (x-1) \, \ln(4) \\
x \, (\ln(3) - \ln(4)) &= - \ln(2^{2}) = - 2 \, \ln(2) \\
x &= - \frac{2 \, \ln(2)}{\ln\left(\frac{3}{4}\right)}
\end{align}
Converting from natural logarithm to a base 10 logarithm: 
Use 
$$\log_{b}(x) = \frac{\log_{d}(x)}{\log_{d}(b)}$$
to convert the base of the logarithms. Here, let $b = e$ and $d = 10$ to yield
\begin{align}
x = - 2 \, \cdot \frac{\log_{10}(2)}{\log_{10}(e)} \, \frac{\log_{10}(e)}{\log_{10}\left(\frac{3}{4}\right)} = -2 \, \frac{\log_{10}(2)}{\log_{10}\left(\frac{3}{4}\right)}.
\end{align}
Strictly solving by base 10 logarithm:
\begin{align}
\log_{10}(3^{y}) &= \log_{10}(4^{y-1}) \\
y \, \log_{10}(3) &= y \, \log_{10}(4) - \log_{10}(4) \\
y \, (\log_{10}(3) - \log_{10}(4)) &= - \log_{10}(4) \\
y &= \frac{- \, \log_{10}(4)}{\log_{10}(3) - \log_{10}(4)} = \frac{2 \, \log_{10}(2)}{\log_{10}\left(\frac{4}{3}\right)}.
\end{align}  
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