## Trending News

Promoted

### 4 Answers

Relevance

- 2 years agoFavorite Answer
Do you know how to solve quadratics? Let e^(x) = u

u^2 - 2u - 24 = 0

u^2 - 2u = 24

u^2 - 2u + 1 = 24 + 1

(u - 1)^2 = 25

u - 1 = -5 , 5

u = 1 - 5 , 1 + 5

u = -4 , 6

e^(x) = -4 , 6

There's no real solution for e^(x) = -4. That's extraneous

e^(x) = 6

x = ln(6)

- 2 years ago
Think of a logarithm as a reverse exponent. So log(10) 100 = 2 could be rewritten as 10^2 = 100.

Still have questions? Get your answers by asking now.