e^2x+e^x-20=0 Find the exact value of x ???

Answer 1

e^(2x) + e^(x) - 20 = 0
Let y = e^(x)
y ² + y - 20 = 0
( y + 5 ) ( y - 4 ) = 0
y = - 5 , y = 4
e^(x) = - 5 , e^(x) = 4
Accepting + ve value of e^(x) = 4
ln e^(x) = ln 4
x ln e = ln 4
x = ln 4

Answer 2

Let u = e^x, the equation becomes:

u^2 + u - 20 = 0

Factor it -
(u + 5) (u - 4) = 0
u = -5, u = 4

e^x = -5 can't be as the range of an exponential functions is greater than 0,

e^x = 4, converting it to logarithmic form, we have x = ln 4

Answer 3

e^x = u

u^2 + u - 20 = 0

(u-4)(u+5) = 0

u = 4 or -5

e^x = 4 or -5

Exclude -5 since e is positive. A positive # raised to any real power is positive or 0.

e^x = 4

ln 4 = x

x = 1.386294361119062...

I can't give you an exact value because x is irrational.

Answer 4

Let y = e^x. Then this equation is:

y^2 + y - 20 = 0

This solves as y = -5 or y=4.

That means that e^x = 4 (e^x = -5 is not possible)
So x = ln(4)

Answer 5

assume you meant

e^(2x) + e^x - 20 = 0

rewrite the expression
(e^x)^2 + e^x - 20 = 0

let y be e^x
y^2 + y - 20 = 0

factor
(y - 4) (y + 5) = 0

y = 4 or -5

recall that y = e^x
4 = e^x
x = ln4

-5 = e^x
x = no root

so the only solution is ln4