Question

(i) If $x={\left(\frac{3}{2}\right)}^2\times {\left(\frac{2}{3}\right)}^{-4}$, find the value of x⁻².
(ii) If $x={\left(\frac{4}{5}\right)}^{-2}÷{\left(\frac{1}{4}\right)}^2$, find the value of x⁻¹.

Answer 1

(i) First, we have to find x.

$$\begin{gathered} x={\left(\frac{3}{2}\right)}^2\times {\left(\frac{2}{3}\right)}^{-4} \\ ={\left(\frac{3}{2}\right)}^2\times {\left(\frac{3}{2}\right)}^4 \\ ={\left(\frac{3}{2}\right)}^6 \end{gathered}$$ --->(a⁻¹ = 1/a)


Hence, x⁻² is:

$$\begin{gathered} x^{-2}={\left({\left(\frac{3}{2}\right)}^6\right)}^{-2} \\ ={\left(\frac{3}{2}\right)}^{-{12}^{}} \\ ={\left({\displaystyle \frac{2}{3}}\right)}^{12} \end{gathered}$$ --->(a⁻¹ = 1/a)


(ii) First, we have to find x.
---> ((a/b)ⁿ = (aⁿ)/(bⁿ))


---> (a⁰ = 1)
Hence, the value of x⁻¹ is:

--->(a⁻¹ = 1/a)
--->(a⁻¹ = 1/a)