# How do you solve 5^(x-1)=3x?

Jul 31, 2017

See below.

#### Explanation:

Making ${e}^{\lambda} = 5$ we have

${5}^{-} 1 {e}^{\lambda x} = 3 x$ or

${5}^{-} \frac{1}{3} = x {e}^{- \lambda x} = \frac{1}{\lambda} \left(\lambda x\right) {e}^{- \lambda x}$ then

$\frac{{5}^{-} 1 \lambda}{3} = - \left(- \lambda x\right) {e}^{- \lambda x}$ and now using the Lambert function

https://en.wikipedia.org/wiki/Lambert_W_function

applying $Y = X {e}^{X} \Leftrightarrow X = W \left(Y\right)$ we have

$- \lambda x = W \left(\frac{- {5}^{-} 1 \lambda}{3}\right)$ or

$x = - \frac{1}{\lambda} W \left(\frac{- {5}^{-} 1 \lambda}{3}\right)$ with $\lambda = {\log}_{e} 5$ or

$x = - \frac{W \left(- {\log}_{e} \frac{5}{15}\right)}{\log} _ e 5$

The Lambert function furnishes two solutions

$x = \left\{0.0752499466729842 , 2.161545847967751\right\}$