we'll use the same log cancellation rule, since this is pretty much the same thing as the other, just recall that ln = logₑ.
[tex] \bf \textit{Logarithm Cancellation Rules}
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log_a a^x = x\qquad \qquad \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{a^{log_a x}=x}
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\textit{logarithm of factors}
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log_a(xy)\implies log_a(x)+log_a(y)
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\rule{34em}{0.25pt}\\\\
ln(x)+ln(5)=2\implies ln(x\cdot 5)=2\implies log_e(5x)=2
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e^{log_e(5x)}=e^2\implies 5x=e^2\implies x=\cfrac{e^2}{5} [/tex]
[tex] \bf \\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
ln(x)-ln(5)=2\implies ln\left( \cfrac{x}{5} \right)=2\implies log_e\left( \cfrac{x}{5} \right)=2\implies e^{log_e\left( \frac{x}{5} \right)}=e^2
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\cfrac{x}{5}=e^2\implies x=5e^2 [/tex]
