Question

Solve the following equation!
10e^x-45e^-x-15=0
Thanks in Advance!

Answer 1

multiply by \(e^x\) and then solve a quadratic equation in \(e^x\)

Answer 2

Could you please show how? :L

Answer 3

\[10e^x-45e^{-x}-15=0\] multiply by \(e^x\) and get
\[10e^{2x}-45-15e^x=0\]

Answer 4

divide by 5 gives
\[2e^{2x}-3e^x-9=0\]
this is a quadratic equation in \(e^x\) as it is the same as
\[2(e^x)^2-3e^x-9=0\]

Answer 5

or you can think of it as letting \(z=e^x\) and you get
\[2z^2-3z-9=0\] if you are lucky it factors

Answer 6

Okay man thank you so much! I never thought of doing it that way!

Answer 7

can you factor this one?
don't forget, that when you solve for \(z\) you then have to write \(z=e^x\) and solve for \(x\), which only requires taking the log

Answer 8

Yeah, or ln, I just needed to get it to where it's a quadratic equation.

Answer 9

oh and throw out the negative solution
one solution is \(z=-\frac{3}{2}\) but \(e^x=-\frac{3}{2}\) is not possible

Answer 10

Yeah that's what I did. There's only 1 solution.

Answer 11

k good!

Answer 12

Thanks alot! You're awesome!