Possible Logarithm Solutions

Since we have $\log_{5} x$ in the equation, our domain is restricted to $x > 0$ . Taking $\log_{5}$ of both sides, we have $\begin{aligned} \log_5 \left(x^{\log_{5} x}\right) &= \log_{5} \left( \frac{x^3}{25} \right) \\ (\log_{5} x) (\log_{5} x) &= 3 \log_{5} x - \log_{5} 25 \\ (\log_{5} x)^2 - 3\log_{5} x + 2 &= 0 \\ (\log_{5}x - 2)(\log_{5}x - 1) &= 0 \\ \end{aligned}$

Therefore $\log_{5}x = 2 \Rightarrow x = 5 ^2 = 25$ and $\log_{5}x = 1 \Rightarrow x = 5$ are solutions. A quick check shows that they are valid solutions. Hence the sum is $5 + 25 = 30$ .

Let $x = 5^{a}$ . The equation becomes $5^{a^{2}} = 5^{a-2}$ .

Taking $log_{5}$ of both sides, we get $a^{2} = a-2$ .

Solving for $a$ , we get $a = 1$ and $a = 2$ as solutions.

Hence, the possible values of $x = 5^{a}$ are $5$ and $25$ .

let log5x be k. then 5^k=x. 5=x^1/k. so the equation can be rewritten as x^k=x^(3-2/k). k=3-2/k; k^2=3k-2; k^2-3k+2=0.
solving the quadratic equation, we get 2 and 1 as values of k. by definition of log, x=5^2=25 and x=5^1=5. 5+25=30