Solving Exponential Equations

**LoveBeachForever** · November 12th 2008, 02:16 PM

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

**Moo** · November 12th 2008, 02:21 PM

Hello,

Originally Posted by LoveBeachForever

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

Let $y=2^x$

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

**Soroban** · November 12th 2008, 03:16 PM

Hello, LoveBeachForever!

Solve: . $\frac{2^x + 2^{-x}}{2}\:=\:3$

Multiply by 2: . $2^x + 2^{-x} \;=\;6$

Multiply by $2^x\!:\;\;2^{2x} + 1 \;=\;6\!\cdot\!2^x \quad\Rightarrow\quad 2^{2x} - 6\!\cdot\!2^x + 1 \:=\:0$

So we have: . $(2^x)^2 - 6(2^x) + 1 \;=\;0$

Let $u \,=\,2^x$ and we have: . $u^2 - 6u + 1 \:=\:0$

The Quadratic Formula gives us: . $u \;=\;3 \pm2\sqrt{2}$

Back-substitute: . $2^x \;=\;3\pm2\sqrt{2}$

Take logs: . $\ln(2^x) \;=\;\ln(3 \pm2\sqrt{2}) \quad\Rightarrow\quad x\ln(2) \;=\;\ln(3 \pm2\sqrt{2})$

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

**LoveBeachForever** · November 12th 2008, 04:30 PM

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

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