# How to solve an equation involving the exponential and the logarithm

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

• 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.wolframalpha.com/input/… Dec 27, 2013 at 11:51
• $x=-1$ is a solution too, but Wolfram alfa doesn't tell you for some reason Dec 27, 2013 at 11:54
• Do you want to know how to compute numerically the other root ? Dec 27, 2013 at 11:59
• Hint: if $f(t)=\log_3\left(\sqrt{t}+1\right)$ then $f^{-1}(t)=\left(3^t-1\right)^2$. Dec 27, 2013 at 12:04
• @Ragnar: Oh, yes. I didn't notice it. Thank you. Dec 27, 2013 at 12:07

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.

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.