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

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

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

• I do nkt think the tag ordinary differential equations have any relevance – user665856 Apr 25 at 18:22
• Why should there be a process? It is equivalent to solving $1^y+3^y=4^y$ where $y=2x$ and there is clearly only one solution which happens to be $y=1$. But there is not an obvious solution to $1^y+3^y=5^y$ (about $0.727$) – Henry Apr 25 at 18:22
• $$x=2$$ is the only real solution. – Dr. Sonnhard Graubner Apr 25 at 18:25
• how does one solve such equation? I want to know the steps of solving such an equation. I have tried using log, I got the wrong answer.(x=0) – Mosiur Rahman Apr 25 at 18:26

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.

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.