Solving $2^x = \sqrt{3^x}+1$ using logarithm. I have problem solving the equation
$$2^x = \sqrt{3^x} +1$$
for $x$ using logarithm. I know the only answer is $2$ which can be proven using graphs or derivatives,etc or by dividing the two sides by $2^x$ which gives the sum of $\sin 60°$ to the power of $x$ and $\cos 60°$ to the power of $x$ equal to $1$, concluding $x=2$.
I'm looking for a way to solve it using logarithm which is not easy because of the "$1$" in one side of the equation.

Here is my approach.We will solve the equation for $x>0$. The case $x \leq 0$  obviously gives us no solution because:
$$ 2^{-x}=\frac{1}{2^x} \leq 1  < 1 + 3^{-\frac{x}{2}}. $$
Then for the case $x > 0$. We define:
$$f(x):= 2^x - 3^{\frac{x}{2}} - 1, ~x \in \mathbb{R}$$
Hence: $$\frac{df}{dx} = \ln 2 \cdot 2^x - \frac{\ln 3}{2}\cdot 3^{\frac{x}{2}}=\frac{1}{2}\cdot \left(\ln 4 \cdot 2^x - \ln 3\cdot \sqrt{3}^{~x} \right)>0, \forall x >  0$$
Hence $f(x)$ is increasing for all $x>0$  and therefore $f(x)$ meet x-axis at only one point and that is where $f(x)=0$. Easily find that $x=2$ satisfy $f(x)=0$ and that is our only solution.
