1. ## Solving Exponential Equations

$\displaystyle \frac{2^x + 2^(-x)}{2}=3$
I honestly have no clue how to do this because I was absent when we went over it. Please help!

2. Hello,
Originally Posted by LoveBeachForever
$\displaystyle \frac{2^x + 2^(-x)}{2}=3$
I honestly have no clue how to do this because I was absent when we went over it. Please help!
Let $\displaystyle y=2^x$

and remember that $\displaystyle a^{-b}=\frac{1}{a^b}$
hence $\displaystyle 2^{-x}=\frac{1}{2^x}$

3. 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: .$\displaystyle x \;=\;\frac{\ln(3 \pm2\sqrt{2})}{\ln(2)} \;\approx\; \pm 2.543106606$

4. ## Thanks!!

I thank both of you so much, I understand this now!