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(a) Given that $\log _{10} p=a, \log _{10} q=b$ and $\log _{10} s=c$, express $\log _{10}\left(\frac{p^{\frac{1}{3}} q^{4}}{s^{2}}\right.$ in terms of $a, b$ and $c$.

(b) The radius of a circle is $6 \mathrm{~cm}$. If the area is increasing at the rate of $20 \mathrm{~cm}^{2} s^{-1}$, find, leaving the answer in terms of $\pi$, the rate at which the radius is increasing.
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\begin{aligned} &(a) \log _{10} p=a ; \log _{10} q=b ; \log _{10} s=c \\ &\log _{10}\left(\frac{p^{\frac{1}{3}} q^{4}}{s^{2}}=\log _{10} p^{\frac{1}{3}}+\log _{10} q^{4}-\log _{10} s^{2}\right. \\ &=\frac{1}{3} \log _{10} p+4 \log _{10} q-2 \log _{10} s \\ &=\frac{1}{3} a+4 b-2 c \end{aligned}
(b) Area of circle $=\pi r^{2}$
Given $r=6 \mathrm{~cm}$
\begin{aligned} &\frac{\mathrm{d} A}{\mathrm{~d} r}=2 \pi r \\ &\frac{\mathrm{d} A}{\mathrm{~d} t}=\frac{\mathrm{d} A}{\mathrm{~d} r} \times \frac{\mathrm{d} r}{\mathrm{~d} t} \\ &20=2 \pi(6) \times \frac{\mathrm{d} r}{\mathrm{~d} t} \\ &\frac{\mathrm{d} r}{\mathrm{~d} t}=\frac{20}{12 \pi} \\ &=\frac{5}{3 \pi} \end{aligned}
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