Given that \[3(4^p) = 7(6^q)\] and \[3(8^{p+1}) = 25(6^q),\] prove that \[2^p = \frac{25}{56},\] and express \(6^q\) also as the quotient of two integers. Determine, to two significant figures, the value of \(p\).

We can rewrite the second equation as \[3 \times 8 \times 8^p = 25 \times 6^q\] and then rewrite \(8^p\) as \(4^p \times 2^p\), which gives us \[3 \times 8 \times 4^p \times 2^p= 25 \times 6^q.\]

Dividing this by the first equation gives \[8 \times 2^p = \frac{25}{7}.\] Therefore \(2^p = \dfrac{25}{56}\), as required.

To find an expression for \(6^q\), we can start by rewriting the \(4^p\) in the first equation in terms of \(2^p\) and substitute the value we’ve just found for \(2^p\), so \(4^p = 2^{2p} = (2^p)^2\).

Therefore the first equation can be rewritten as \[3 \times (2^p)^2 = 7 \times 6^q.\] Substituting the value \(\frac{25}{56}\) for \(2^p\) gives \[7 \times 6^q = 3 \times \Bigl(\frac{25}{56}\Bigr)^2\] and therefore \[6^q = \frac{3 \times 25^2}{7 \times 56^2} = \frac{1875}{21952}.\]

Finally, calculating \(p\): \[\begin{equation} \label{eq:log} 2^p = \frac{25}{56} \end{equation}\]so we can take logs to any base and rearrange to find \(p\). \[p\log 2 = \log\frac{25}{56}\] and therefore, dividing both sides by \(\log 2\), \[p = -1.2 \quad \text{to 2 s.f.}\]

Here we can use log to any base, say base 10. We also need a calculator to work out the numerical value of \(p\).

If your calculator can calculate logs to any base of your choosing, then you can solve equation \(\eqref{eq:log}\) immediately by calculating \(\log_2 \left(\dfrac{25}{56}\right)\).