Ask Uncle Colin: Some Messy Powers

Dear Uncle Colin,

I’m told that $\frac{16^p \times 8^q}{4^{p+q}}=2^n$, and I need to find $n$ in terms of $p$ and $q$. How would I do that?

$q$ Uppishly In $n$ Equation

Hi, $q$UI$n$E1, and thanks for your message!

There are several ways to approach this, as per always. Let’s start with a couple of power-manipulation tricks, and end up with logs.

Powers of 2

The one that jumps out at me is that all of the numbers involves are powers of 2. That means we can rewrite the left-hand-side as:

$\frac{\left(2^4\right)^p \times \left(2^3\right)^q}{\left(2^2\right)^{p+q}}$

Or as $\frac{2^{4p + 3q}}{2^{2p+2q}}$

Then we can simply subtract the powers to get $2^{2p+q}$, and state that $n= 2p+q$.

Splitting things up

Alternatively, we can split the denominator up as $4^p \times 4^q$.

Then the left-hand-side becomes $\frac{16^p \times 8^q}{4^p \times 4^q}$, or $4^p \times 2^q$.

Again using the fact that $4^p = 2^{2p}$, we get $2^{2p+q}$.

Logarithms

We could, sledgehammer-wise, take logs:

$p \ln(16) + q \ln(8) - (p+q)\ln(4) = n \ln(2)$

Dividing everything by $\ln(2)$ - possibly using your calculator if the Mathematical Ninja is safely in another time-zone, gives:

$4p + 3q - 2(p+q) = n$

Then it’s simply $2p + q = n$ again.


Hope that helps!

- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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