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I am trying to solve for $x$ in the equation $\log_4(2x) = \frac{1}{2}x^2 - 1$. I have tried converting the logarithmic expression to exponential form, but I am not able to isolate $x$ in the resulting equation.

This is what I have tried as of now:

$$\log_4(2x) + \log_4(4) = \frac{1}{2}x^2$$

$$\log_4(8x) = \frac{1}{2}x^2$$

$$2\log_4(8x) = x^2$$

$$\log_4(64x^2) = x^2$$

$$64x^2 = 4^{x^2}$$

after which I am not too sure on how to find x

Tuvasbien
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Thamognya Kodi
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  • It doesn't look like an equation you can solve without a computer. You will probably need some numerical method to find an approximate solution. – Arnaldo Feb 06 '23 at 13:54
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    Taking the square root in a couple places and working backwards gives $\sqrt{\log_2 (8x)}=x$ which is all the simplification I believe is possible without resorting to the Lambert $W$ function... – abiessu Feb 06 '23 at 13:55
  • Is there a reason why? any pattern that shows that it needs to be solved with an computer. – Thamognya Kodi Feb 06 '23 at 13:55
  • In general almost any equation you write up won’t have a solution you can compute by hand, so the question is not why but why not. – Vivaan Daga Feb 06 '23 at 13:56
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    If you draw the graph of those two functions...then you'll see two points of intersection...one if them is two as we can easily solve for $64x^2=4^{x^2}$ ...it will give $x=\pm2$ but negative solution is not valid.....now the other point of intersection has to be numerically evaluated – MathStackexchangeIsNotSoBad Feb 06 '23 at 13:57
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    $x=a^x$ is a typical equation that is known to almost always require numerical methods... – abiessu Feb 06 '23 at 13:57
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    Do not worry about the special function. You will learn about it quite soon and I hope and wish that you will enjoy it (I felt in love with its 64 years ago). It is very use. On the search bar, just ype Lambert and notice 4158 entries. Cheers :-) – Claude Leibovici Feb 06 '23 at 14:51
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    the reason why solving by rearranging with elementary functions isn't possible: see https://math.stackexchange.com/questions/4586412/how-can-we-show-that-az-ez-and-a-ln-z-z-have-no-elementary-inverse – IV_ Feb 06 '23 at 17:30

2 Answers2

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For your curiosity.

You want to find the zeros of function $$f(x)=\frac{x^2}{2}-\frac{\log (2 x)}{\log (4)}-1$$ By insepection $x=2$ is a root

The first derivative $$f'(x)=x-\frac{1}{x \log (4)}$$ cancels at $x=\frac{1}{\sqrt{\log (4)}}$ and the second derivative test shows that this is a maximum.

Since $x >0$, rewrite $$\log(2x)=\log(2)+\frac 12 \log(x^2)$$ and let $t=x^2$ to face $$g(t)=2t \log(2)-\log(t)-6\log(2)$$ Th only explicit solution involves Lambert function $$t=-\frac{1}{2 \log (2)}W\left(-\frac{\log (2)}{32}\right)$$ Since the argument is small, use the series expansion (it is given in the linked page) $$W(y)=y-y^2+\frac{3 y^3}{2}+O\left(y^4\right)$$ which will give $$t \sim \frac{1}{64}+\frac{\log (2)}{2048}+\frac{3 \log ^2(2)}{131072}=0.0159744 \quad \implies \quad x=0.126390$$ while the exact solution is $x=0.126392 $.

Claude Leibovici
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$$\log_4{(2x)}=\frac{x^2}{2}-1$$

As stated already in a previous answer, try $x=2$: $$x=2: \log_4{(2\cdot 2)}=\frac{2^2}{2}-1\Leftrightarrow 1=1$$

So $x=2$ is a solution. So far nothing new (from previous answer).

Next I would transform the equation into an exponential one: $$\begin{align}\log_4{(2x)}&=\frac{1}{2}(1+\log_2 x)\\ \log_2 x&=x^2-3\\8x&=2^{x^2}\\f(x)&=2^{x^2}-8x\\f(0)&\gt 0, f(1)\lt 0\\f(0)&\gt 0, f(\frac{1}{2})\lt 0,…f(\frac{1}{8})\gt 0, f(\frac{3}{16})\lt 0\end{align}$$

I would continue to chase the root by halving the interval. So far I could tell that the root is located between $\frac{1}{8}$ and $\frac{3}{16}$. this plot was obtained using demos.com With red is $y=2^{x^2}$ and with blue is $y=8x$. The plot was obtained from desmos.com

WindSoul
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