Question

Solve 6e^(x)-4e^(-x)=5 for x

Answer 1

$6e^x-4e^{-x}=5\\\\6e^x-\dfrac{4}{e^x}=5$

Substitute $e^x=t\ \textgreater \ 0$ and

$6t-\dfrac{4}{t}=5\quad|\cdot t\\\\\\6t^2-4=5t\\\\6t^2-5t-4=0\\\\\\a=6\qquad b=-5\qquad c=-4\\\\\\\Delta=b^2-4ac=(-5)^2-4\cdot6\cdot(-4)=25+96=121\\\\\\\sqrt{\Delta}=\sqrt{121}=11\\\\\\ t_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-(-5)-11}{2\cdot6}=\dfrac{-6}{12}=-\dfrac{1}{2}\ \textless \ 0\\\\\\ t_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-(-5)+11}{2\cdot6}=\dfrac{16}{12}=\boxed{\dfrac{4}{3}\ \textgreater \ 0}$

so

$t=\dfrac{4}{3}\\\\\\e^x=\dfrac{4}{3}\quad|\ln(\ldots)\\\\\\\ln(e^x)=\ln\left(\dfrac{4}{3}\right)\\\\\\x=\ln(4)-\ln(3)\\\\x=\ln(2^2)-\ln(3)\\\\\boxed{x=2\ln(2)-\ln(3)}$

ln(x) is the natural logarithm.