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Algebra
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solve for x by changing to exponential form. 5 ln(x)=9
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\(\bf 5 ln(x)=9\implies ln(x)=\cfrac{9}{5}\implies log_e(x)=\cfrac{9}{5} \\ \quad \\ \textit{keep in mind that}\quad log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}} \\ \quad \\ log_{\color{red}{ e}}({\color{blue}{ x}})=\cfrac{9}{5}\implies ?\)
so it would be 6.05?
6.05?
ohh ahemm yes
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