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find the exact solution, using common logarithms, and a two-decimal-place approximation of the solution 4^x-3(4^-x)=8

Precalculus
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\[\Large 4^x-3(4^{-x})=8\]first subtract 8 on both sides \[\Large 4^x-3(4^{-x})-8=0\]I would multiply both sides by 4^x \[\Large 4^{2x}-3-8(4^x) = 0\](think about exponent laws and you'll see how to get it. Now it's like a quadratic.
im terribly confused as to how i would apply common logarithms to this problem. :(
You could say: Let 4^x=y then you would have : y^2-8y-3=0 Solve for y using the quadratic formula. Set each answer equal to 4^x and take the log of both sides.

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Other answers:

|dw:1386036292570:dw| i ended up with this
|dw:1386036395791:dw| would it turn to this?
Can't read that. \[\Large 4^{2x}-3-8(4^x) = 0\]if you understand to here, do as merts said. Let y = 4^x \[\Large y^2 -8y - 3 = 0\]solve using quadratic formula.
I ended up with \[4^x= \frac{ 8+\sqrt{52} }{ 2 }\]
The answer i should end up with\[\frac{ \log(4+\sqrt{19)} }{\log4 }\]
Check your math in the quadratic formula, looks like your numbers aren't right.
\[y=\frac{8\pm \sqrt{64-4(1)(-3)}}{2}=\frac{8\pm \sqrt{76}}{2}=\frac{8\pm2\sqrt{19}}{2}=4\pm \sqrt{19}\]
\[4^x=4+\sqrt{19}\] \[x \log_{10}4=\log_{10}(4+\sqrt{19}) \]
Divide both sides by log 4
I am so grateful to both of you. Thank you so much! :)

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