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If log10^2=m and log10^3=n, find log10^24 in term of m and n​

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let's recall that whenever omitted, the base is assumed to be 10.


\begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}


\log_(10)(2)=m\implies \log(2)=m\hspace{5em}\log_(10)(3)=n\implies \log(3)=n \\\\[-0.35em] ~\dotfill\\\\ \log_(10)(24)\implies \log(24)\implies \log(8\cdot 3)\implies \log(2^3\cdot 3) \\\\\\ \log(2^3)+\log(3)\implies 3\log(2)+\log(3)\implies 3m+n

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