Question

Solve for x7D%20" id="TexFormula1" title="0 = - { e }^{ - x} + 3 {e}^{3x} " alt="0 = - { e }^{ - x} + 3 {e}^{3x} " align="absmiddle" class="latex-formula">

Answer 1

$\bf \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{log_a a^x = x}\qquad \qquad a^{log_a x}=x~\hfill\stackrel{recall}{ln=log_e}\qquad log_e(e^z)=z \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf 0=-e^{-x}+3e^{3x}\implies e^{-x}=3e^{3x}\implies \cfrac{1}{e^x}=3e^{3x}\implies 1=e^x\cdot 3e^{3x} \\\\\\ 1=3e^xe^{3x}\implies \cfrac{1}{3}=e^{x+3x}\implies \cfrac{1}{3}=e^{4x}\implies ln\left( \cfrac{1}{3} \right)=ln\left( e^{4x} \right) \\\\\\ ln\left( \cfrac{1}{3} \right)=4x\implies \cfrac{ln\left( \frac{1}{3} \right)}{4}=x$

and you plug that in your calculator to get about -0.27465307216702742285.