$\displaystyle
\frac{2^x + 2^(-x)}{2}=3
$
I honestly have no clue how to do this because I was absent when we went over it. Please help!
Hello, LoveBeachForever!
Solve: .$\displaystyle \frac{2^x + 2^{-x}}{2}\:=\:3$
Multiply by 2: .$\displaystyle 2^x + 2^{-x} \;=\;6$
Multiply by $\displaystyle 2^x\!:\;\;2^{2x} + 1 \;=\;6\!\cdot\!2^x \quad\Rightarrow\quad 2^{2x} - 6\!\cdot\!2^x + 1 \:=\:0$
So we have: .$\displaystyle (2^x)^2 - 6(2^x) + 1 \;=\;0$
Let $\displaystyle u \,=\,2^x$ and we have: .$\displaystyle u^2 - 6u + 1 \:=\:0$
The Quadratic Formula gives us: .$\displaystyle u \;=\;3 \pm2\sqrt{2}$
Back-substitute: .$\displaystyle 2^x \;=\;3\pm2\sqrt{2} $
Take logs: .$\displaystyle \ln(2^x) \;=\;\ln(3 \pm2\sqrt{2}) \quad\Rightarrow\quad x\ln(2) \;=\;\ln(3 \pm2\sqrt{2})$
Therefore: .$\displaystyle x \;=\;\frac{\ln(3 \pm2\sqrt{2})}{\ln(2)} \;\approx\; \pm 2.543106606$