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We are currently discussing logarithms and exponential equations. I am currently answering a problem set until I stumbled upon this question:

$2(2^{2x})=4x+64$

I tried using the usual methods such as log and ln but I could not get past the $4x+64$. The best I was able to do was:

$(2x+1)\log (2) = \log (4x+64)$

If I tried solving the left side, I have:

$0.602x+0.3010 = \log(4x+64)$

If this is the right way, then I am stuck in this part of the equation. If I made a mistake in any of these steps, please let me know and please tell me how I can solve this question correctly. Thank you.

Lapnesnoj
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  • I assume you mean $2 \cdot 2^{2x} = 4x + 64$? WolframAlpha says the solution isn't nice at all. – VTand Nov 06 '21 at 14:57
  • It can be solved using Lambert w function https://en.m.wikipedia.org/wiki/Lambert_W_function – RAHUL Nov 06 '21 at 15:00
  • @VTand yes I tried searching the solution online and it was not pretty. – Lapnesnoj Nov 06 '21 at 15:02
  • @RAHUL I was wondering if there was a solution without using Lambert W function, since we have not tackled that. – Lapnesnoj Nov 06 '21 at 15:03
  • If you do not want to use the Lambert-W function, then the only way possible is to find an approximation of the solution using the secant or Newton or any other known algorithm. Note: you could simplify your equation by dividing both sides by 4, but this will not help. – MasB Nov 06 '21 at 15:10

2 Answers2

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That equation can only be solved using Lambert $W$ function. $$ 2^{2x}=2x+32 $$ $$ (2x+32)2^{-2x} = 1 $$ $$ -(2x+32)2^{-2x-32}=-2^{-32} $$ We know that we can represent $2$ as $e^{\ln2}$: $$ -(2x+32)e^{-(2x+32)\ln2}=-2^{-32}\ \big |\cdot \ln 2 $$ $$ -(2x+32)\ln2\ e^{-(2x+32)\ln2}=-2^{-32}\ln 2 $$ Knowing that $W(xe^x)=x$: $$ W(-(2x+32)\ln2\ e^{-(2x+32)\ln2})= -(2x+32)\ln2=W(-2^{-32}\ln2) $$ Now it's easy to show that: $$ x = -\cfrac{W(-2^{-32}\ln2)+32\ln2}{2\ln2} $$

Techno TCHe
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If you want to find the positive zero of function $$f(x)=2^{2x}-(2x+32)$$ only Lambert function can provide an explicit solution. Otherwise, numerical methods.

$f(x)$ is highly nonlinear. Consider instead function $$g(x)=\log(2^{2x})-\log(2x+32)$$ which is close to linearity. Expand it as a Taylor series around $x=0$ $$g(x)=-5\log (2)+x \left(2 \log (2)-\frac{1}{16}\right)+\frac{x^2}{512}+O\left(x^3\right)$$ Solve the quadratic equation to get, as an estimate, $$x_0=\frac{1}{2} \left(32-1024 \log (2)+\sqrt{10240 \log (2)+(1024 \log (2)-32)^2}\right)=2.60800$$ while the solution is $x=2.60896$.

For sure, we can do better $$g(x)=-\log \left(\frac{9}{4}\right)+ \left(\log (4)-\frac{1}{18}\right)(x-2)+\frac{1}{648} (x-2)^2+O\left((x-2)^3\right)$$ Solve the quadratic equation to get, as another estimate, $x=2.60895$.

Claude Leibovici
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