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

anyone help!!

Printable View

- May 8th 2010, 07:39 AMmastermin346indices problem
solve the equation $\displaystyle 8^{2x-3}=\frac{1}{\sqrt{4^x+2}}$

anyone help!! - May 8th 2010, 08:06 AMwonderboy1953First step
Square both sides to get rid of the radical.

- May 8th 2010, 04:41 PMmastermin346
dont get the answer

- May 8th 2010, 04:48 PMAnonymous1
$\displaystyle 8^{2x-3}=\frac{1}{\sqrt{4^x+2}}$

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

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

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

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)