# indices problem

• May 8th 2010, 07:39 AM
mastermin346
indices problem
solve the equation $\displaystyle 8^{2x-3}=\frac{1}{\sqrt{4^x+2}}$

anyone help!!
• May 8th 2010, 08:06 AM
wonderboy1953
First step
Square both sides to get rid of the radical.
• May 8th 2010, 04:41 PM
mastermin346
• May 8th 2010, 04:48 PM
Anonymous1
Quote:

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.