how many answers do this equation have$3^{2x}-34(15^{x-1})+5^{2x}=0$ How many answers do this equation have?
$3^{2x}-34(15^{x-1})+5^{2x}=0$
My Attempt:$3^{2x}+5^{2x}=34(15^{x-1})$.Now what to do?

Letting $A = 3^x$ and $B = 5^x$, we get:
$$A^2 - \dfrac{34}{15}AB + B^2 = 0.$$
Since this equation is symmetric in $A$ and $B$, we can consider it as a quadratic in the variable $A$.
Using the quadratic formula:
$$A = \dfrac{\dfrac{34}{15}B \pm \sqrt{\left(\dfrac{34}{15}B\right)^2 - 4B^2}}{2} = \dfrac{\dfrac{34}{15}B \pm \sqrt{\left(\dfrac{16}{15}B\right)^2}}{2} = \dfrac{\dfrac{34}{15}B \pm \dfrac{16}{15}B}{2},$$
from which we obtain
$$A_1 = \dfrac{\dfrac{34}{15}B - \dfrac{16}{15}B}{2} = \dfrac{9}{15}B = \dfrac{3}{5}B,$$
and
$$A_2 = \dfrac{\dfrac{34}{15}B + \dfrac{16}{15}B}{2} = \dfrac{25}{15}B = \dfrac{5}{3}B.$$
Substituting back $A = 3^x$ and $B = 5^x$, we obtain:
$$3^x = \left(\dfrac{3}{5}\right){5^x},$$
or
$$3^x = \left(\dfrac{5}{3}\right){5^x}.$$
From the first equation, we have:
$$3^{x-1} = 5^{x-1}$$
while from the second equation, we get:
$$3^{x+1} = 5^{x+1}.$$
Since $\gcd(3,5) = 1$, then we have either
$$3^{x-1} = 1$$
or
$$3^{x+1} = 1$$
(i.e., either $x = 1$ or $x = -1$).

Conclusion: Since $x = 1$ or $x = -1$ satisfy the original equation, then we have obtained TWO SOLUTIONS.

