# Ask Uncle Colin: Lost in Logarithms

Dear Uncle Colin,

I’m a bit lost trying to solve $3\ln(x+2) - \ln(x^2 + x - 2) + \ln\left(\frac{x-1}{16}\right)=0$. How would you go about it?

- Logarithmic Opportunities Shine Through

Hi, LOST, and thanks for your message!

I would start by splitting the last log into its component parts: $\ln\left(\frac{x-1}{16}\right) \equiv \ln(x-1) - \ln(16)$.

I’d then move the negative logs to the other side ((Yeah, yeah, I know I’m technically adding the same thing to both sides. I don’t care.))

- $3 \ln(x+2) + \ln(x-1) = \ln(x^2+x-2) + \ln(16)$

Getting rid of the logs gives:

- $(x+2)^3(x-1) = 16(x^2+x-2)$

The RHS factorises:

- $(x+2)^3(x-1) = 16(x+2)(x-1)$

Now, $x$ cannot be -2 or 1, because they’re not in the original domain. That’s good: it means I can divide by them and rearrange to give:

- $(x+2)^2 = 16$

That means $x+2$ is either 4 or -4, and that $x=-6$ or $x=2$.

However, it can’t be -6, because that would break the original logarithms – the solution is thus $x=2$.

### An interesting error

In solving this, I originally made a subtle and possibly interesting error.

I spotted that $x^2 + x -2$ factorised as $(x+2)(x-1)$ and concluded, incorrectly, that $\ln(x^2 + x - 2) \equiv \ln(x+2) + \ln(x-1)$.

At least, it’s not true *for all possible values of $x$*. It’s fine if $x > 1$, so that both of the RHS logarithms are defined. However, when $x<-2$, the LHS is defined and the RHS isn’t.

It’s not even the same as $\ln | x+2 | + \ln | x-1 | $ – that’s defined for $-2 < x < 1$, where the LHS isn’t. What a mess! |

In fact, $\ln(x^2+x-2)$ is the same as $\ln(x+2)+\ln(x-1)$ if $x>-1$ and $\ln(-2-x) + \ln(1-x)$ for $x < -2$.

For *this* question, though, we can assert that $x>1$ to ensure that the last logarithm is valid, in which case the interesting error isn’t an error. We work as follows:

- $3\ln(x+2) - \ln(x+2) - \ln(x-1) + \ln(x-1) - \ln(16) = 0$
- $2\ln(x+2) = \ln(16)$
- $\ln(x+2) = \ln(4)$
- $x=2$.

Hope that helps!

- Uncle Colin