Logarhythm

For $\log_3 (x^2-729)$ to be a whole number, $x^2-729$ must be a power of 3. Let $x^2-729 = 3^n$ , where $n$ is a whole number. Then

$\begin{aligned} x^2 & = 3^n + 729 \\ & = 3^n + 3^6 \\ & = 3^6(3^{n-6}+1) \\ \implies x^2 & = 27^2(3^{n-6}+1) \end{aligned}$

Since the LHS $x^2$ is a perfect square, $3^{n-6}+1$ must also be a perfect square. The first solution is when $n-6=1$ or $n=7$ , then $3^1 + 1 = 4 = 2^2$ , a perfect square. We find that there is no solution for $n=8$ and $n=9$ . This means that the next acceptable $x > 27 \sqrt{3^3+1} > 100$ . Hence there is only one solution, that is when $n=7$ , $\implies x= 27 \times 2 = \boxed{54}$ .

For $\log_3(x^2-729)$ to be a whole number, $x^2-729 = (x+27)(x-27)$ must be a power of three. Since $x>0$ , $x+27>0$ and, therefore, $x-27>0$ as well (otherwise the expression would be undefined). This means that both $x+27$ and $x-27$ are powers of three. Say $x+27=3^n$ and $x-27=3^m$ (note $m<n$ ). Then $54 = 3^n-3^m = 3^m(3^{n-m}+1)$ . Since $3^{n-m}+1$ cannot be divisible by three, $3^m$ is the largest power of three which divides 54, namely $27$ . Thus $m=3$ and, consequently, $x=\boxed{54}$ is the only solution.

Note: the stipulation that $x<100$ is unnecessary. The only positive integer solution for $x$ is $x=54$ . There is a negative solution: $x=-54$ .

$x^2=3^y+3^6$ , where $y$ has to be a whole number.

$x=3^3 \sqrt {3^{t-6}+1}$

Only t=7 makes it a perfect square, hence $x =27√4=54$