1. ## Solve for x

$\displaystyle \frac{1}{2^x}-\frac{1}{x^2}=0$

$\displaystyle \frac{x^2-2^x}{2^x(x^2)}=0$

I dont know what to do from here.

2. Originally Posted by JimmyRP
$\displaystyle \frac{1}{2^x}-\frac{1}{x^2}=0$

$\displaystyle \frac{x^2-2^x}{2^x(x^2)}=0$

I dont know what to do from here.
two "obvious" solutions are x = 2 and x = 4 ... there is one more that is negative, but you'll have to use a calculator to find it.

3. Originally Posted by skeeter
two "obvious" solutions are x = 2 and x = 4 ... there is one more that is negative, but you'll have to use a calculator to find it.
could you expand on how you would get the negative. I want to learn I don't want just the answers :P

4. Originally Posted by JimmyRP
could you expand on how you would get the negative. I want to learn I don't want just the answers :P
you would have to graph the function $\displaystyle x^2 - 2^x$ with a calculator and find the zeros.

note that the negative value cannot been found using elementary algebraic techniques.

5. could you point me to the forum where i could learn these techniques?