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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.

Arctic Char
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asdpb
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    Why do you expect logarithms will actually help here? As you say, "because of the '1' in one side of the equation" logarithms don't actually make any sense to use. Do not make the mistake of assuming that just because a question is easy to write that it is easy to answer. Do not make the mistake of assuming that every equation has an elementary solution. The general solution to $a^x = b^x + c$ is very likely going to be non-elementary, maybe involving non-elementary things like the Lambert W function if such a solution is even named. – JMoravitz Sep 04 '20 at 19:18
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    The best you may hope to be able to do for the general case is to use numerical methods such as Newton's Method to get better and better approximations to the solution. – JMoravitz Sep 04 '20 at 19:21
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    See answers in https://math.stackexchange.com/questions/3810472/find-all-x-in-mathbbr-such-that-left-sqrt2-sqrt2-rightx-left#comment7855267_3810472. I think all answers there can be modified to answer your question. – Batominovski Sep 04 '20 at 21:22

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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.

Nguyễn Quân
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