Oxford University MAT 2019 Solved Papers

The equations are equivalent to;
$\mathit{b}={\mathit{a}}^{\mathit{c}},\mathit{a}={\mathit{b}}^{\mathit{c}+3∕2},\mathit{a}={\mathit{c}}^{\mathit{b}}$
Raise each side of the second equation to the power of $\mathit{c}$. Then $\mathit{b}={\mathit{b}}^{{\mathit{c}}^2+3\mathit{c}∕2}$, so we must have $1={\mathit{c}}^2+3\mathit{c}∕2$ (since $\mathit{b}>1$ for the logarithms to make sense). This has solutions $\mathit{c}=-2$ or $\mathit{c}=-1∕2$, but we know $\mathit{c}>0$ so $\mathit{c}=1∕2$. The remaining equations read $\mathit{a}={\mathit{b}}^2$ and $\mathit{a}=2^{-\mathit{b}}$. Sketch both curves; There is exactly one positive solution. The answer is (a)