Thread: Using natural logs to solve for x

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.

Or, more directly, take the logarithm of both sides of the original equation:
$\displaystyle ln(3e^{-x})= -x+ ln(3)= ln(4^{3x-1})= (3x-1)ln(4)$

Now you have -x+ ln(3)= 3ln(4)x- ln(4) which is a linear equation.