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![\bf \textit{Logarithm Change of Base Rule} \\\\ \log_a b\implies \cfrac{\log_c b}{\log_c a}\qquad \qquad c= \begin{array}{llll} \textit{common base for }\\ \textit{numerator and}\\ denominator \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 3=5^{x+1}\implies \log_7(3)=\log_7\left( 5^{x+1} \right)\implies \log_7(3)=(x+1)\log_7(5) \\\\\\ \cfrac{\log_7(3)}{\log_7(5)}=x+1\implies \log_5(3)=x+1\implies \log_5(3)-1=x\implies -0.3174 \approx x](/tpl/images/0173/5549/44eb9.png)
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Solve 3=5^(x+1) using change of base...
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