# Math

Solve for x, 6e^(x) - 4e^(-x)=5

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1. 6 e^x - 4/e^x = 5
let z = e^x
6 z -4/z = 5
6 z -4/z - 5 = 0
6 z^2 -5 z - 4 = 0 this will have 2 roots, we may only want one of them
(3z-4)(2z+1) = 0
z = 4/3 or -1/2
if e^x = 4/3
6(4/3) -4/(4/3) = 22/3 - 3 which is not 5
if e^x = -1/2
-6/2 - 4/(-1/2) = -3 + 8 = 5
but there is no such thing, can not take ln of neg number
ln e^x = x = ln(-.5)

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2. 6e^(x) - 4e^(-x)=5
To make things more readable, let u = e^x. Now we have
6u - 4/u = 5
6u^2 - 4 = 5u
6u^2 -5u - 4 = 0
(3u-4)(2u+1) = 0
u = 4 or -1/2
Looks like there are two solutions, but
e^x is never negative, making the only real solution
e^x = 4/3
x = ln(4/3)

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3. if e^x = 4/3
6(4/3) -4/(4/3) = 8 - 3 which IS 5 whew

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