Solving for X that is an exponent? Do have to use logs?

I'm practicing some exponent problems out of perhaps the worst precalculus book in the world -- Advanced Mathematics by Richard G. Brown -- and I just can't see to get this one. The question simply states to "Solve."

2^x + 2^(-x) =5/2

The answer is 1 and -1 but I just can't seem to figure out the work. I've been using Precalculus Solved to find out how certain questions are worked out, but this question prompted the software to give me an error.

If some could help me, I'd really appreciate it.

Thanks in advanced~

Re: Solving for X that is an exponent? Do have to use logs?

Try multiplying both sides through by $\displaystyle 2^x$

Re: Solving for X that is an exponent? Do have to use logs?

$\displaystyle 2^x+\frac{1}{2^x}=\frac{5}{2}$

$\displaystyle \frac{2^{2x}+1}{2^x}=\frac{5}{2}$

$\displaystyle 2\left[2^{2x}+1\right]=5\left[2^x\right]$

This is a quadratic equation in disguise.

$\displaystyle 2y^2-5y+2=0$

$\displaystyle (2y-1)(y-2)=0$

Re: Solving for X that is an exponent? Do have to use logs?

Holy Batman! Thanks for the help! Did not see that... Thank again... I can take a break now!