Solve for x: 3e^(-x) = 4^(3x-1)
Thanks
$\displaystyle 3e^{(-x)} = 4^{(3x-1)}$
$\displaystyle \dfrac{3}{e^x}=4^{3x-1}$
$\displaystyle 3 = e^{(x)}\cdot 4^{3x-1}$
take log on both sides:
$\displaystyle \log{(3)} = \log\bigg(e^{(x)}\cdot 4^{3x-1}\bigg)$
now use $\displaystyle \log(AB)=\log(A)+\log(B)$ on the RHS.