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If In (50) = ln(2) + k ln(5), then k =

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\begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} \\\\[-0.35em] ~\dotfill\\\\ \ln(50)=\ln(2) + k\ln(5)\implies \ln(50)=\ln(2) + \ln(5^k)\implies \ln(50)=\ln(2\cdot 5^k) \\\\\\ 50=2\cdot 5^k\implies \cfrac{50}{2}=5^k\implies 25=5^k\implies 5^2=5^k\implies 2=k

User Gustavo Sousa
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