Question

How do I solve ln(x) + ln(5) = 2 ?

Answer 1

we'll use the same log cancellation rule, since this is pretty much the same thing as the other, just recall that ln = logₑ.

`\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{a^(log_a x)=x} \\\\\\ \textit{logarithm of factors} \\\\ log_a(xy)\implies log_a(x)+log_a(y) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(x)+ln(5)=2\implies ln(x\cdot 5)=2\implies log_e(5x)=2 \\\\\\ e^(log_e(5x))=e^2\implies 5x=e^2\implies x=\cfrac{e^2}{5}`

`\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( (x)/(5) \right)}=e^2 \\\\\\ \cfrac{x}{5}=e^2\implies x=5e^2`