indices problem

**mastermin346** · May 8th 2010, 07:39 AM

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

anyone help!!

**wonderboy1953** · May 8th 2010, 08:06 AM

Square both sides to get rid of the radical.

**mastermin346** · May 8th 2010, 04:41 PM

dont get the answer

**Anonymous1** · May 8th 2010, 04:48 PM

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.

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