Respuesta :

[tex]\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad a^{log_a x}=x\qquad \leftarrow \textit{we'll use this one} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ u=\ln(x)\implies u=\log_e(x)\implies e^u=e^{\log_e(x)}\implies \boxed{e^u=x} \\\\[-0.35em] ~\dotfill\\\\ v=\ln(y)\implies v=\log_e(y)\implies e^v=e^{\log_e(y)}\implies \boxed{e^v=y} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{y^7}{x^5}\implies \cfrac{(e^v)^7}{(e^u)^5}\implies \cfrac{e^{7v}}{e^{5u}}\implies e^{7v}\cdot e^{-5u}\implies e^{7v-5u}[/tex]

Answer:

e^7v-5u

Step-by-step explanation:

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