# Math Help - Solving exponential equations.

1. ## Solving exponential equations.

Can someone please show me how to solve this?

e^x - e^-x = 1
2

Also my professor told me to multiply each side by e^x.. can someone help? Thanks.

so I've tried it and solved it and got x=2.41 and x= -.41 (which when I plug it in, it doesn't equal 1).

e^X (e^x - e^-x) = 2e^x
which then results to (Let A = e^X)

A^2 - 2A -1 = 0.

2. Originally Posted by EliteNewbz
Can someone please show me how to solve this?

e^x - e^-x = 1
2

Also my professor told me to multiply each side by e^x.. can someone help? Thanks.

so I've tried it and solved it and got A=2.41 and A= -.41 (which when I plug it in, it doesn't equal 1).

e^X (e^x - e^-x) = 2e^x
which then results to (Let A = e^X)

A^2 - 2A -1 = 0.
Since $e^x>0$ for all $x\in \mathbb{R}$ the variable A must be greater than zero. So the you can't use the negative solution.

Solve for x: $e^x=1+\sqrt{2}~\implies~x\approx 0.88137...$

3. Originally Posted by EliteNewbz
Can someone please show me how to solve this?

e^x - e^-x = 1
2

Also my professor told me to multiply each side by e^x.. can someone help? Thanks.

so I've tried it and solved it and got x=2.41 and x= -.41 (which when I plug it in, it doesn't equal 1).

e^X (e^x - e^-x) = 2e^x
which then results to (Let A = e^X)

A^2 - 2A -1 = 0.
Solve the quadratic for A. As $b^2-4ac = 8$ there are two real solutions in A

$A = \frac{2+2\sqrt{2}}{2} = 1+\sqrt{2}$
$A = 1-\sqrt{2}$

$e^x = 1+\sqrt{2} \: \rightarrow \: x = ln(1+\sqrt2)$

$e^x = 1-\sqrt{2} \: \rightarrow x = ln(1-\sqrt2) = ln(\sqrt{2}-1)+i\pi$

4. Thank you! I must have done something wrong then!