Question

How do i solve e^(x) ? 20e^(?x) ? 1 = 0?

Thankyou for your time in helping me solve this question!

Answer

e^(x) - 20e^(-x) - 1 = 0

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

(e^(x) - 5)(e^(x) + 4) =0

e^(x) = 5 --> x = ln(5)

e^(x) = -4 --> not possible

Jul 3 at 19:49

with math skills.

Jul 3 at 23:35

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

(e^x - 5)(e^x + 4) = 0

--->x = ln 5

Jul 4 at 3:44

y - 20/y - 1 = 0

y 2 - 20 - y = 0

y 2 - y - 20 = 0

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

y = 5 is acceptable

e^x = 5

x = ln 5

Jul 4 at 8:16

Multiply through by e^x

(e^x)2 – (e^x) – 20 = 0. This is quadratic in e^x

(e^x – 5)(e^x + 4) = 0; e^x = –4, 5

e^x = –4 has no solution since logs of negative numbers are undefined

ln(e^x) = x = ln(5)

Jul 4 at 13:11