<s>[INST] Express $\log_5 288$ in terms of decimal logarithms $\log 2$ and $\log 3$ Assuming $a=\log 2$ and $b=\log 3$ (log is the base 10 logarithm). I have to find $\log_5 288$. How can I do this?
I've tried transforming $\log2$ to $\frac{\log_5 2}{\log_5 10}$ and same for $b$.  Then it's $$\frac{5\log_5 2+2\log_5 3}{\log_5 10}=\frac{\log_5 288}{\log_5 10}=5\log2+2\log5=5a+3b$$ Is that correct?
 [/INST] \begin{align}
\log_5 288 & = \frac{\log_{10} 288}{\log_{10} 5} = \frac{\log_{10}(2\cdot2\cdot2\cdot2\cdot2\cdot3\cdot3)}{\log_{10} (10/2)} \\[15pt]
& = \frac{\log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}3 + \log_{10}3 +  }{\log_{10} 10 - \log_{10} 2} \\[15pt]
& = \frac{\log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}3 + \log_{10}3 +  }{1 - \log_{10} 2}
\end{align}
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