# Variable exponent solve for x

How can I solve this exponent problem using simple math only?

We need to solve for $x$

$2^{2x}-(3.2)^{x+2} + 32 = 0$

The second term here is $3.2$ not $3\cdot2$ ie 3 decimal 2 not 3 into 2.

My attempt

$2^{2x}- \dfrac{32^{x+2}}{10^{x+2}} = -32$

$\dfrac{\left(10^{x+2}\times2^{2x}\right)- {2^{5x+10}}}{10^{x+2}} = -32$

I cant solve it further :(

## EDIT

http://www.wolframalpha.com/input/?i=%282%29%5E%282x%29+-+%283.2%29%5E%28x%2B2%29+%2B+32+%3D+0

Wolfram alpha told that its quite complex.

If you look at the plot of the function, you would notice that the first solution is close to $1$. So, let us use Newton method which, starting from a "reasonable" guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, for the problem (I let you computing the derivative), starting at $x_0=1$, the successive itererates will be $1.09924$, $1.09409$, $1.09408$ which is the solution for six significant figures.
If you do the same for the second solution, starting with $x_0=11$, the successive itererates will be $10.6685$, $10.4827$, $10.4290$, $10.4251$ which is the solution for six significant figures.
It seems impossible that, in precalculus, they gave you this equation to solve. More than likely, there is a typo somewhere and the problem was to find the solution of $$2^{2x}-3\times 2^{x+2} + 32 = 0$$ Just as Chinny84 answered, rewrite it as $$2^{2x}-3\times 2^{x}2^2 + 32 =2^{2x}-12\times 2^x+32= 0$$ and set $y=2^x$. The equation then reduces to a quadratic $y^2-12y+32=0$ the roots of which being $4$ and $8$; from this, compute the corresponding $x$'s.
• You still don't know but you will very soon ! I gave that to you but I am ready to bet that the equation you need to solve is $2^{2x}-3\times 2^{x+2} + 32 = 0$ which will make full sense for precalculus. To me,just as for Chinny84 probably, there is a typo somewhere. Cheers :-) – Claude Leibovici Mar 9 '15 at 11:06