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
On
$$\log_4{(2x)}=\frac{x^2}{2}-1$$
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
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 $.