<s>[INST] 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?
 [/INST] Substitute $a=3^x$ and $b=5^x$ to get
$a^2-\frac{34}{15}ab+b^2=0$
Multiply both sides by $15$ to get,
$15a^2-34ab+15b^2=0 \Rightarrow 15a^2-25ab-9ab+15b^2=0 \Rightarrow 5a(3a-5b)-3b(5a-3b)=0 \Rightarrow (3a-5b)(5a-3b)=0$
Now, let $\frac{a}{b}=t$
Dividing both sides by $b^2=5^{2x} \neq 0$, we get 
$t=\frac{3}{5}$ or $t=\frac{5}{3}$
Back-substituting, we get $t=\frac{3^x}{5^x}=(\frac{3}{5})^x$
So we are just left to check two cases


*

*$(\frac{3}{5})^x=\frac{3}{5}$. Dividing both sides by $\frac{3}{5}$ we get,  $(\frac{3}{5})^{x-1}=1 \Rightarrow x-1=0 \Rightarrow x=1$,  as $a^x=1 \Rightarrow x=0$, if $a \neq 0$

*$(\frac{3}{5})^x=\frac{5}{3}=(\frac{3}{5})^{-1} $. Multipying both sides by $\frac{3}{5}$ we get,  $(\frac{3}{5})^{x+1}=1 \Rightarrow x+1=0 \Rightarrow x=-1$,  as $a^x=1 \Rightarrow x=0$, if $a \neq 0$

Thus, $x= \pm 1$ and there are two solutions.

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