The equation $\log_3(\sqrt{x+1}+1)=(3^{x+1}-1)^2\,$ has two solutions, but I can't solve the equation.
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Are you sure that this equation has two solutions? Wolfram alpha tells that there is only one solution. Of course, I don't know if it is true.http://www.wolframalpha.com/input/?i=log_3%28%28x%2B1%29%5E%281%2F2%29%2B1%29%3D%283%5E%28x%2B1%29-1%29%5E2 – mathlove Dec 27 '13 at 11:51
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1$x=-1$ is a solution too, but Wolfram alfa doesn't tell you for some reason – Ragnar Dec 27 '13 at 11:54
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Do you want to know how to compute numerically the other root ? – Claude Leibovici Dec 27 '13 at 11:59
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6Hint: if $f(t)=\log_3\left(\sqrt{t}+1\right)$ then $f^{-1}(t)=\left(3^t-1\right)^2$. – Constructor Dec 27 '13 at 12:04
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@Ragnar: Oh, yes. I didn't notice it. Thank you. – mathlove Dec 27 '13 at 12:07
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After the nice remark from Constructor and the simpler formulation given by labuwx, let see how to solve the equation
3^t - Sqrt[t] - 1 = 0
I think that Newton iterative scheme is very simple. Starting with a guess (say t_old), the iterates are given by
t_new = t_old - f(t_old) / f'(t_old)
In your case, f(t) = 3^t - Sqrt[t] - 1 and labuwx showed that there is a root close to 0.5. So, start the iterations with t_old = 0.5. According to Newton scheme, the successive iterates will then be 0.479139, 0.478604 which is the solution.
For sure, you could do the same with the original equation.
Claude Leibovici
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As Constructor wrote the equation is: $f(t)=f^{-1}(t)$ where $t=x+1$ and $f(t)=log_3(\sqrt{t}+1)$
If there is such $t_0$ that $f(t_0)=f^{-1}(t_0)$ <=> $f(t_0)=t_0 => log_3(\sqrt{t_0}+1)=t_0 => 3^{t_0}=\sqrt{t_0}+1$
$3^t$ is convex, $\sqrt{t}+1$ is concave so cannot be more than 2 solutions: $t_1=0$, $t_2$ should be $1-0.5213960...$ accordint to WolframAlpha, but I don't know how to find a closed form (if exists).
$t_1=0 => x_1=-1$ which is the beginning of the domain of $log_3(\sqrt{x+1}+1)$ so maybe that is why WolframAlpha didn't find it numerically.
