Can we solve these simultaneous exponential equations?

Can we solve these simultaneous exponential equations?

Question

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 $p$.

Solution

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

$p$

:

$$\begin{equation} 2^p = \frac{25}{56} \end{equation}$$

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 $(log)$ immediately by calculating $\log_2 \left(\dfrac{25}{56}\right)$.