what is the solution to this equation: $1 + 3^{x/2} = 2^x$?

Question

what is the solution to this equation: $1 + 3^{x/2} = 2^x$ ?

The answer is $x = 2$. I want to know the process.

Answer 1

First look at negative $x$. You have that $\log(1 + 3^{x/2}) >0$ whereas $\log(2^x) <0$ so there cannot be a solution.

For positive $x$, note that $3^{x/2}$ grows less than $2^{x}$ for all positive $x$, which can simply be shown by differentiating. Further note that $1 + 3^{x/2}$ and $2^{x}$ are strictly monotonously rising functions.

Since for $x=0$, we have that $1 + 3^{x/2} >2^{x}$, and for sufficiently large $x$, we have that $1 + 3^{x/2} < 2^{x}$, there will be exactly one solution (with positive $x$) for $1 + 3^{x/2} = 2^{x}$. As others have already noted, you cannot directly compute this solution. However, if you have found $x=2$, you are done.

Answer 2

Iterate it: We rearrange to $$x=\log(1+\sqrt{3^x})$$

Then apply $$x_{n+1}=\log(1+\sqrt{3^{x_n}}); x_0=1$$

You can see from a calculator that $$\lim_{t\to\infty}x_t=2$$

Unfortunately, we cannot solve these.