What is the solution to 2^x + 2^-x = 3?

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The Attempt at a Solution



2^x + 2^-x = 3

2^x + (1 / ((2^x)) = 3

(4^((x^2)) +1) / (2^x) = 3

(4^((x^2)) + 1) = 6^ x
 
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No, 2(3^x) is not the same 6^x.
 
Multiply both sides by 2x instead.
 
My bad. I forgot that they must have the same base to multiply them. Also, If you're referring to the last step, I did multiply both sides by 2^x. I'm stuck at that point though.
 
let u = 2^x
solve the quadratic
after solve for u solve for x
 
Thanks that helped a lot. Thanks everyone for the quick replies
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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