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

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

Andreas
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Mosiur Rahman
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  • I do nkt think the tag ordinary differential equations have any relevance –  Apr 25 '19 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 '19 at 18:22
  • $$x=2$$ is the only real solution. – Dr. Sonnhard Graubner Apr 25 '19 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 '19 at 18:26

2 Answers2

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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.

Andreas
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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.

Rhys Hughes
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