Solve the equation in real numbers: $x-6+\frac{2}{\sqrt{x-2}}=\frac{1}{3}\log_3(\frac{x}{x^3+54})$

Question

Solve the equation in real numbers: $x-6+\frac{2}{\sqrt{x-2}}=\frac{1}{3}\log_3(\frac{x}{x^3+54})$

My work: I have managed to find that $3$ is a solution to the problem.I tried to prove that this is the only solution by proving that the LHS will be bigger/smaller than the RHS if $x\neq3$ but I failed to finish it.I was also thinking about a solution using convexity of some function and then proving that there are at most 2 solutions but I couldn't do that either.Can anybody help me,pls?

Answer 1

Setting $\frac{x}{x^{3}+54}=3^{y}$ we have that

$\log_{3}\Big(\frac{x}{x^{3}+54}\Big)= \log_{3}\Big(3^{y}\Big)=y\log_{3}\Big(3\Big)=y$

and therefore

$y=3x-18+\frac{6}{\sqrt{x-2}}$

This function has a minimal value in the point where the first derivative is null.

$\frac{dy}{dx}=3-\frac{3}{(x-2)^{3/2}}=0$

$(x-2)^{3/2}=1$

By squaring both members, we get:

$(x-2)^{3}=1$

$x^{3}-6x^{2}+12x-9=0$

$(x-3)(x^{2}-3x+3)=0$

The real root $x=3$ is the solution.

Verification. Replacing $x=3$ we have

$3-6+\frac{2}{\sqrt{3-2}}=\frac{1}{3}\log_{3}\Big(\frac{3}{3^{3}+54}\Big)$

$-1=\frac{1}{3}\log_{3}\Big(\frac{1}{27}\Big)= \frac{1}{3}\log_{3}\Big(3^{-3}\Big)=-1$.