how many answers do this equation have$3^{2x}-34(15^{x-1})+5^{2x}=0$

Question

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?

Answer 1

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.

Answer 2

$$3^{2x}-34(15^{x-1})+5^{2x}=0$$ $$3^{2x}-\frac{34}{15}(15^{x})+5^{2x}=0$$ $$15\cdot3^{2x}-34\cdot 3^x \cdot 5^x+15\cdot5^{2x}=0$$ $$15\cdot\left(\frac{3}{5}\right)^{2x}-34\cdot \left(\frac{3}{5}\right)^{x}+15=0$$ $\left(\frac{3}{5}\right)^{x}=t$ $$15\cdot t^{2}-34\cdot t+15=0$$ $t=\frac35$ or $t=\frac53$

$x=1$ or $x=-1$

Answer 3

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 $x=-1$.

Answer 4

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.