If $\dfrac{3}{2} \text{ log a} + \dfrac{2}{3} \text{ log b } - 1 = 0$, find the value of a⁹.b⁴

Given,

$\Rightarrow \dfrac{3}{2} \text{ log a} + \dfrac{2}{3} \text{ log b } - 1 = 0 \\[1em] \Rightarrow \text{log a}^{\dfrac{3}{2}} + \text{log b}^{\dfrac{2}{3}} = 1 \\[1em] \Rightarrow \text{log a}^{\dfrac{3}{2}} + \text{log b}^{\dfrac{2}{3}} = \text{log 10} \\[1em] \Rightarrow \text{log } (a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}}) = \text{log 10} \\[1em] \Rightarrow (a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}}) = 10$

Cubing and then squaring both sides, we get :

$\Rightarrow [(a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}})^3]^2 = [(10)^3]^2 \\[1em] \Rightarrow (a^{\dfrac{3}{2}} \times b^{\dfrac{2}{3}})^6 = 10^6 \\[1em] \Rightarrow a^{\dfrac{3}{2} \times 6}.b^{\dfrac{2}{3} \times 6} = 10^6 \\[1em] \Rightarrow a^{\dfrac{18}{2}}.b^{\dfrac{12}{3}} = 10^6 \\[1em] \Rightarrow a^9.b^4 = 10^6.$

Hence, a⁹.b⁴ = 10⁶.