(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I have a curiosity question. The precalc textbook asks me to solve by using a graphing calculator but I want to do it algebraically. I know that the equations intersect through the graphing calculator -- one intersection at x = 1.07808, another around -1.5. The two equations are:

[tex]4^x + 4^{-x} = 8 - 2x - x^2[/tex]

2. Relevant equations

3. The attempt at a solution

First I took the log of both sides:

[tex]x\ln{4} + {-x}\ln{4} = \ln{8} - \ln(2x) - \ln(x^2)[/tex]

(not sure how to create a new line with tex, since \\ is not working for me)

[tex]x(\ln{4} - \ln{4}) = \ln{8} - \ln{2} + \ln{x} - \ln{x} - \ln{x}[/tex]

Forgive my mathematical illiteracy. I sort of guessed from that I could cancel the left side since ln(4) - ln(4) = 0, but I'm probably wrong there. And anyway, what if the equation didn't work out that conveniently? If the left side can be canceled and the ln(x)es on the right-side cancel, then it seems like it could become:

[tex]\ln{x} = \ln{8} - \ln{2}[/tex]

which is the same as:

[tex]e^{\ln{8} - \ln{2}} = x = 4[/tex] ??

Forgive my terrible math skills.

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# Solving 4^x + 4^-x = 8 - 2x - x^2

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