# How do you solve 3^x= 3x+5?

Apr 3, 2018

See below.

#### Explanation:

Making $y = 3 x + 5$ we have

${3}^{\frac{y - 5}{3}} = y$ or

${3}^{\frac{y}{3}} {3}^{- \frac{5}{3}} = y$

now making ${3}^{\frac{1}{3}} = {e}^{\lambda}$ we have

${e}^{\lambda y} {3}^{- \frac{5}{3}} = y$ or

${3}^{- \frac{5}{3}} = y {e}^{- \lambda y}$ or

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

At this point we use the Lambert function

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

$Y = X {e}^{X} \Leftrightarrow X = W \left(Y\right)$ and then

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

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

$x = \frac{- 5 L o g 3 - 3 W \left(- \left(L o g \frac{3}{9 \times {3}^{\frac{2}{3}}}\right)\right)}{3 L o g 3}$

We have then two solutions

$x = - 1.60981$ and $x = 2.24048$

Attached a plot showing the intersections of $y = {3}^{x}$ and $y = 3 x + 5$ 