Question:
By what number should $\left(\frac{5}{3}\right)^{-2}$ be multiplied so that the product may be $\left(\frac{7}{3}\right)^{-1} ?$
Solution:
Expressing as a positive exponent, we have:
$\left(\frac{5}{3}\right)^{-2}=\frac{1}{(5 / 3)^{2}} \quad \cdots\left(a^{-1}=1 / a\right)$
$=\frac{1}{25 / 9} \quad \ldots>\left((a / b)^{n}=\left(a^{n}\right) /\left(b^{n}\right)\right)$
$=\frac{9}{25}$
and
(7/3)−1 = 3/7. ---> (a−1 = 1/a)
We have to find a number x such that
$\frac{9}{25} \times x=\frac{3}{7}$
Multiplying both sides by 25/9, we get:
$x=\frac{3}{7} \times \frac{25}{9}=\frac{1}{7} \times \frac{25}{3}=\frac{25}{21}$
Hence, (5/3)−2 should be multiplied by 25/21 to obtain (7/3)−1.