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

1 Answer
Sep 4, 2016

Answer:

#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#