Logarithm in the Exponent

Taking log to the base b, we get $(\log_b a)(\log_b c)=(\log_b c)(\log_b a)$

So this is true for all a,b,c satisfying the given restraints. So the answer is $99^3=970299$

$a^{\log_b c}=c^{\log_b a} \\ \implies a^{\log_b c \times \frac{1}{\log _a b}}=c \\ \implies a^{\log_b c \times {\log _a b}}=c$

$\implies a^{\log_a c}=c$

Let, $\log_a c=p$ then $a^p=c$ which is from above equation.From last equation $c=a^p$ Putting $p$ in the second equation weget the identity:

$a^{\log_a c}=c$ .

So,this is valid for any $(a,b,c)$ So number of solutions are $99 \times 99 \times 99=99^3=970299$

This equality is, indeed, an Identity (Try to prove it). That means, this equality is true for every $a$ , $b$ and $c$ as long as $a, b, c \in R^+-\{1\}$ .

So, there will be $99 \times 99 \times 99 = 970299$ ordered triples of positive integers $(a,b,c)$ , where $99 = 100-2+1$ .