0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ I do nkt think the tag ordinary differential equations have any relevance $\endgroup$ – user665856 Apr 25 at 18:22
  • $\begingroup$ 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 $0.727$) $\endgroup$ – Henry Apr 25 at 18:22
  • $\begingroup$ $$x=2$$ is the only real solution. $\endgroup$ – Dr. Sonnhard Graubner Apr 25 at 18:25
  • $\begingroup$ 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) $\endgroup$ – Mosiur Rahman Apr 25 at 18:26
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.