A question relating to exponent laws

Question

Here is the question:

If:

$$125*(3^x) = 27*(5^x)$$

Then find the value of $x$.

Here is what I have done so far:

I found that I can manipulate the numerical values to get the same bases:

$\to 125 = 5^3$ and $27 = 3^3$

$\to 5^{3-x} = 3^{3-x}$

Here is where I got stuck. I do not have similar bases anymore so where can I go from here? Should I change it to a logarithm or should I do a trial and error to see what value(s) work or is there a way to get one of the sides into the same base as the other?

Answer 1

You have $5^{3-x}=3^{3-x}$. Divide both sides by $3^{3-x}$, then $\left(\frac{5}{3}\right)^{3-x}=1\Rightarrow 3-x=0\Rightarrow x =3$.

(Remember $\frac{5^{3-x}}{3^{3-x}}=\left(\frac{5}{3}\right)^{3-x}$ using the index law $\frac{a^m}{b^m}=\left(\frac{a}{b}\right)^{m}$.)

Answer 2

$$\frac{3^{3-x}}{5^{3-x}} = 1 \implies \left(\frac{3}{5} \right)^{3-x} = 1$$

and you know that $a^0 = 1$ for any real number $a$.