Question 4Given $\dfrac{\log a}{p}=\dfrac{\log b}{q}=\dfrac{\log c}{r}=\log x,$ all logarithms to the same base and $x\ne1.$ If $\dfrac{b^2}{ac}=x^y,$ then $y$ is:$\dfrac{q^2}{p+r}$$\dfrac{p+r}{2q}$$2q-p-r$$2q-pr$$q^2-pr$