# Thread: Solve equation

1. ## Solve equation

Hello,

How do I solve the following equation? $\displaystyle 2^2^x-3*2^x-40 = 0$

I tried the following, but apparently this is not correct, i.e. score = 0%.

$\displaystyle 2^2^x-3*2^x-40 = 0$
$\displaystyle 2^2^x-\sqrt{3}*2^2^x = 40$
$\displaystyle (1-\sqrt{3})*2^2^x = 40$
$\displaystyle 2^2^x = \frac{40}{1-\sqrt{3}}$
$\displaystyle 2x = \log_{2}\frac{40}{1-\sqrt{3}}$
$\displaystyle x = \frac{\log_{2}\frac{40}{1-\sqrt{3}}}{2}$

2. ## Re: Solve equation

You changed 3 to root3 Why?
Let y=2^x Then 2^2x= (2^x)^2=y^2
So equation becomes y^2-3y-40=0
(y-8)(y+5)=0 y=8 0r-5
So 2^x=8 giving x=3 ( and 2^x=-5 has no solution)
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