Originally Posted by
mastermin346 solve the equation $\displaystyle 8^{2x-3}=\frac{1}{\sqrt{4^x+2}}$
anyone help!!
$\displaystyle 8^{2x-3}=\frac{1}{\sqrt{4^x+2}}$
$\displaystyle \implies 8^{2x-3}=(4^x+2)^{-\frac{1}{2}}$
$\displaystyle \implies 8^{-4x+6}=4^x+2$
$\displaystyle \implies 8^{-4x+6}-4^x = 2$
Now, convince your self that $\displaystyle 8^n = 4^{\frac{3}{2}n}$ $\displaystyle \forall$ $\displaystyle n.$
$\displaystyle \implies 4^{\frac{3}{2}(-4x+6)}-4^x = 2$
$\displaystyle \implies 4^{(-6x+9)}-4^x = 2$
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.