How to solve exponential expressions?

Question

There are some questions that I have.

Question 1) $$ (x^2/5)^3 = 2^6/5^y$$ To find the $y$ I used the same base

$$ 1/5^3 = 1/5y$$ Teacher told that the exponent will be the same if equaled, so $ y = 3$. What my question is how would I find $x$?? Because I can't find the same base.

Question 2)

$$ (5 \cdot 6 )^2 = 5^x6^y$$ They don't have the same base, how would I solve them?

Answer 1

HINT (answer to the first question, the 2th you'll find out):

$$\left(\frac{x^2}{5}\right)^3=\frac{2^6}{5^y} \Longleftrightarrow$$ $$\frac{x^6}{125}=\frac{64}{5^y} \Longleftrightarrow$$ $$8000=x^6 \cdot 5^y \Longleftrightarrow$$ $$8000=x^6 \cdot 5^y \Longleftrightarrow$$ $$x^6 \cdot 5^y = 8000 \Longleftrightarrow$$ $$ 5^y = \frac{8000}{x^6} \Longleftrightarrow$$ $$ y = \frac{\ln\left(\frac{8000}{x^6}\right)}{\ln(5)}+\frac{2i\pi n}{\ln(5)}$$

With $n\in\mathbb{Z}$

Answer 2

I am not sure $$ \left(\frac{x^2}{5}\right)^3 = \frac{x^6}{5^3} = \frac{2^6}{5^y} $$ so the simplest solution is $y=3$ and $x=2$. similarly for the second equation. $$ (5\cdot 6)^2 = 5^2\cdot 6^2 = 5^x6^y $$ implies $x=2$ and $y=2$.

But your question is not entirely clear to me.

Answer 3

i would write $$\left(\frac{x^2}{5}\right)^3=\frac{2^6}{5^y}$$ is equivalent to $$\left(\frac{x}{2}\right)^6=5^{3-y}$$
for 2) we can write $$5^2\cdot 6^2=5^x\cdot 6^y$$ and then $$5^{2-x}=6^{y-2}$$ and we can solve this for one variable.