# Math Help - indices problem

1. ## indices problem

solve the equation $8^{2x-3}=\frac{1}{\sqrt{4^x+2}}$

anyone help!!

2. ## First step

Square both sides to get rid of the radical.

4. Originally Posted by mastermin346
solve the equation $8^{2x-3}=\frac{1}{\sqrt{4^x+2}}$

anyone help!!
$8^{2x-3}=\frac{1}{\sqrt{4^x+2}}$

$\implies 8^{2x-3}=(4^x+2)^{-\frac{1}{2}}$

$\implies 8^{-4x+6}=4^x+2$

$\implies 8^{-4x+6}-4^x = 2$

Now, convince your self that $8^n = 4^{\frac{3}{2}n}$ $\forall$ $n.$

$\implies 4^{\frac{3}{2}(-4x+6)}-4^x = 2$

$\implies 4^{(-6x+9)}-4^x = 2$

Hmmmm, well here is the answer...

Code:
>> syms x;
>> solve('8^(2*x-3) = (4^x + 2)^(-1/2)')

ans =

((log(2)*log(2, z))/2 + pi*i*k)/log(2)
...and it's an ugly complex number, what do you know.