How to solve an equation involving the exponential and the logarithm

Question

The equation $\log_3(\sqrt{x+1}+1)=(3^{x+1}-1)^2\,$ has two solutions, but I can't solve the equation.

Answer 1

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.

Answer 2

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.