Question

Pre calc

Solve e^2x +3e^x- 28=0

Answer 1

`{e}^(x) = y \\ {y}^(2) + 3y - 28 = 0 \\ (y - 4)(y + 7) = 0 \\ y = 4 \\ y = - 7 \\ {e}^(x) = 4 \\ x = ln(4) \\ x = ln( - 7)`

Answer 2

x=ln(4)

Explanation:

e^2x +3e^x- 28=0

u^2 + 3u - 28 = 0 ==> substitute u for e^x

u^2 - 4u + 7u - 28 = 0

u(u-4) + 7(u-4) = 0

(u-4)(u+7)=0 ==> simplify

u-4=0 u+7=0

u=4 u=-7

e^x=4 e^x=-7 ==> substitute e^x for u

ln(e^x)=ln(4) ln(e^x)=-7 ==> take natural log on each side

x=ln(4) x=ln(-7), which is not possible since natural log>0

x=ln(4)