If a² = log x, b³ = log y and $\dfrac{a^2}{2} - \dfrac{b^3}{3}$ = log c, find c in terms of x and y.

Given,

a² = log x and b³ = log y

Substituting value of a² and b³ in $\dfrac{a^2}{2} - \dfrac{b^3}{3}$ = log c, we get :

$\Rightarrow \dfrac{a^2}{2} - \dfrac{b^3}{3} = \text{log c} \\[1em] \Rightarrow \dfrac{\text{log x}}{2} - \dfrac{\text{log y}}{3} = \text{log c} \\[1em] \Rightarrow \dfrac{\text{3 log x - 2 log y}}{6} = \text{log c} \\[1em] \Rightarrow \text{log x}^3 - \text{log y}^2 = \text{6 log c} \\[1em] \Rightarrow \text{log x}^3 - \text{log y}^2 = \text{log c}^6 \\[1em] \Rightarrow \text{log c}^6 = log \dfrac{x^3}{y^2} \\[1em] \Rightarrow c^6 = \dfrac{x^3}{y^2} \\[1em] \Rightarrow c = \sqrt[6]{\dfrac{x^3}{y^2}}.$

Hence, c = $\sqrt[6]{\dfrac{x^3}{y^2}}.$