Question

solve for x using log
2^x+2^-x=3

Answer 1

\[2^x + 2^{-x} = 3\]
i remember seeing a problem like this before..

you know how to do this right @KingGeorge

Answer 2

Ah one of these. First thing to do, is convert it to a quadratic.

Answer 3

\[2^x+2^{-x}=3 \implies (2^x)^2-3\cdot2^x+1=0\]Note that this is a quadratic in \(2^x\). I don't believe this factors nicely, so now we use the quadratic formula.

Answer 4

\[a=1\]\[b=-3\]\[c=1\]\[2^x=\frac{3\pm\sqrt{9-4}}{2}\]\[2^x=\frac{3\pm\sqrt{5}}{2}\]From here, look at \[2^x=\frac{3+\sqrt{5}}{2}\]\[2^x=\frac{3-\sqrt5}{2}\]and take log base 2 of both sides.

Answer 5

thanks! how did you convert it to (2 x ) 2 −3⋅2 x +1=0 though?

Answer 6

I multiplied by \(2^x\). So \[2^x\left(2^x+2^{-x}\right)=2^x\cdot3\]And then distribute and subtract \(3\cdot2^x\) from both sides.