Question
How do i solve e^(x) ? 20e^(?x) ? 1 = 0?
Jul 4 at 13:11
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
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