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?
 A: 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.

A: $$3^{2x}-34(15^{x-1})+5^{2x}=0$$
$$3^{2x}-\frac{34}{15}(15^{x})+5^{2x}=0$$
$$15\cdot3^{2x}-34\cdot 3^x \cdot 5^x+15\cdot5^{2x}=0$$
$$15\cdot\left(\frac{3}{5}\right)^{2x}-34\cdot \left(\frac{3}{5}\right)^{x}+15=0$$
$\left(\frac{3}{5}\right)^{x}=t$
$$15\cdot t^{2}-34\cdot t+15=0$$
$t=\frac35$ or $t=\frac53$
$x=1$ or $x=-1$
A: Divide the equation by $5^{2x}$. Then, you have 
$$\left(\left(\dfrac35\right)^x\right)^2-\frac{34}{15}\left(\dfrac35\right)^x+1=0$$
Let $a=\left(\frac35\right)^x$. Then, our equation becomes 
$$a^2-\frac{34}{15}a+1=0$$
or 
$$15a^2-34a+15=0$$
or
$$(5a-3)(3a-5)=0$$
So, $a=5/3$ or $a=3/5$, thus, $x=1$ or $x=-1$.
A: 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.

