Solve $\sqrt x = 1 + \ln(3 + x)$ algebraically

Question

I am having trouble with this homework problem. I am able to graph and find the solution, but I am curious as to how one would do this algebraically.

The way I began, was subtracting $1$ on both sides and taking the $\exp$ of both sides to rid of $\ln$. I am not sure how to proceed.

Answer 1

There are a lot of equations that you can write down that we don't know how to solve in exact form, like $x + e^x = 2$, $\cos(x) = x$, etc. I believe this is one of them. You might be able to solve it if you allow yourself to use the Lambert-W function, a special function designed to be used in this sort of situation. I haven't worked with $W$ that much, so I am not sure. But this simply replaces your question with the question of "How do I compute $W$?" Some software packages could calculate it, but in practice, it's probably easier to just use numerical methods for solving equations like you have already done on your calculator.

My guess is that this problem was specifically designed to not be solvable in exact form, so that you can practice using your calculator. In real-world applications many equations are not explicitly solvable. Besides, the class of equations that you can solve in "exact form" is pretty artificial, since you're allowed to use some kinds of functions ($\log$, $\sin$, etc.) but not others ($W$).

Answer 2

Differentiating $\sqrt{x} - (1+\log(3+x))$, we obtain $$\frac{1}{2 \sqrt{x}} - \frac{1}{3+x}$$ which is always positive because we rearrange to $$\frac{3+x-2 \sqrt{x}}{2 (3+x) \sqrt{x}}$$ but $3+x > 2 \sqrt{x}$ always (by squaring).

Therefore there is at most one solution; there is indeed a solution because evaluating at $1$ yields $1$ on the LHS and $1+3 \log{4} > 1$ on the RHS; while evaluating at $100$ gives $10$ on the LHS and $1+\log(103) \simeq 5.6$ on the RHS. So by the intermediate value theorem, the unique root is somewhere between $1$ and $100$.

That's all I've got, I'm afraid.