<s>[INST] Solving for x in logarithmic equation $\log_4(2x) = \frac{1}{2}x^2 - 1$ 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
 [/INST] $$\log_4{(2x)}=\frac{x^2}{2}-1$$
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 $\frac{3}{16}$.

With red is $y=2^{x^2}$ and with blue is $y=8x$. The plot was obtained from desmos.com
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