Ask Uncle Colin: An Uncommon Logarithm

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 $\log ((x+3{)}^3)$ - 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