*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I'm a bit stumped by a logs question with a variable base: $\log_{\sqrt[3]{x+3}}(x^3 + 10x^2 + 31x + 30) = 9$. I know the basics of logarithms, but this is currently beyond me.

-- Obtaining Underwhelming Grade, Having To Review Every Definition

Hello, OUGHTRED, and thanks for your message! That *is* a bit of a monster, but it does fall apart nicely when you apply the basics (in fact, I'd say the algebra element was more challenging than the logarithms bit!)

Another way of writing $\log_a(b)$ is $\frac{\log(b)}{\log(a)}$, so let's do that first:

$\frac{\log(x^3 + 10x^2 + 31x + 30)}{\log(\sqrt[3]{x+3})} = 9$

Using the power law for logs, we can rewrite $\log(\sqrt[3]{x+3})$ as $\frac{1}{3}\log(x+3)$, and then cross-multiply:

$\log(x^3 + 10x^2 + 31x + 30) = 9 \times \frac{1}{3} \log(x+3)$

That right hand side is $3 \log(x+3)$, or $\log \left((x+3)^3 \right)$ - and now we have a plain logarithm on both sides. We can remove those to leave us with:

$x^3 + 10x^2 + 31x + 30 = (x+3)^3 = x^3 + 9x^2 + 27x + 27$

Now it's just algebra! That reduces to $x^2 + 4x + 3 = 0$, and factorises as $(x+3)(x+1)=0$.

So, either $x=-3$ or $x=-1$... or does it?

It can't possibly be the case that $x=-3$, because then we'd have a logarithm with base 0, and that's Not Allowed - the original left hand side isn't defined there. However, $x=-1$ works perfectly well. Moral of the story: always check your answers work!

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.