#### Answer

a) $2 \thinspace species$
b) $333.58 \thinspace m^2$

#### Work Step by Step

a) We are given $S=0.7A^{0.3}$ and $A=60\thinspace m^2$. If we plug in $A$ into the equation, we get $S\approx2.39$. Since we can't have 2.39 species, we'll round down to $2 \thinspace species$.
b) We are given $S=4$.
$4=0.7A^{0.3}$
$\frac{4}{0.7}=A^{0.3}$
$\log{\frac{4}{0.7}}=log{A^{0.3}}$
Using log properties, we can bring down the $0.3$.
$\log{\frac{4}{0.7}}=0.3log{A}$
$\frac{1}{0.3}\log{\frac{4}{0.7}}=log{A}$
$A=10^{\frac{1}{0.3}\log{\frac{4}{0.7}}}$
$A=10^{\frac{10}{3}\log{\frac{40}{7}}}$
$A=(10^{\log{\frac{40}{7}}})^{\frac{10}{3}}$
$A=(\frac{40}{7})^{\frac{10}{3}}\approx333.58\thinspace m^2$