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 :(



Wolfram alpha told that its quite complex.


I do not think that there is a closed form of the solution and, probably, numerical methods would need to be used to solve the problem.

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.

  • $\begingroup$ I don't know what you are talking about, but this seems correct. $\endgroup$ – Steve Gates Mar 9 '15 at 11:04
  • 2
    $\begingroup$ 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 :-) $\endgroup$ – Claude Leibovici Mar 9 '15 at 11:06

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