Review question

Can we solve these simultaneous exponential equations? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5940

Solution

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 $p$: $$$\label{eq:log} 2^p = \frac{25}{56}$$$

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 $\log_2 \left(\dfrac{25}{56}\right)$.