Question

How do you solve `e^x=2`?

Answer 1

`x=ln2`

Explanation:

Make use of the `color(blue)"law of logarithms"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(logx^n=nlogx)color(white)(a/a)|)))`

Using this law allows us to obtain x as a multiplier

First step - take ln of both sides.

`rArrlne^x=ln2rArrxcancel(lne)^1=ln2rArrx=ln2`