$\displaystyle \frac{1}{2^x}-\frac{1}{x^2}=0$ $\displaystyle \frac{x^2-2^x}{2^x(x^2)}=0 $ I dont know what to do from here.
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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.
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
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.
could you point me to the forum where i could learn these techniques?
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