# Ask Uncle Colin: Are the log laws... lacking?

Dear Uncle Colin,

I have an equation to solve: $\ln(x^2) = 2 \ln(4)\, x \ne 0$.

I tried to solve it by applying the log laws: $2 \ln(x) = 2 \ln(4)$, so $x=4$.

However, a bit of thought shows that $x=-4$ is also a solution – but that doesn’t seem to come out of the laws!

My eyes have gone all wibbly. Do I need to see an optician?-- When Are Logarithmic Laws Insufficiently Strict?

Hi, there, WALLIS – what a great question! I would probably have approached the question in just the same way as you and made the same error. I, thanks to corrective surgery, have 20-20 vision, so I don’t think you need to go to Specsavers on account of this problem.

The problem is quite a subtle one – and it’s to do with the domain of the function. Normally, an expression with a logarithm in it is defined for $x > 0$ rather than $x \ne 0$ – for obvious reasons: if you stick a negative number into a logarithm, bad things happen.

Here, the clue is that it’s defined for $x \ne 0$ – meaning that negative $x$ is fair game. This is ok, because we’re squaring the $x$ to make it positive before we put it into the log.

The log laws aren’t *wrong*, as such, but they’re incomplete: they should probably be written as $\log_a\left(x^n\right) = n \log_a \left(\left|x\right|\right)$ for $x \ne 0$ (although then you need to be careful about fractional powers when $x < 0$, which is another problem for another day).

Another way to approach this is to draw a graph: $y=\ln\left(x^2\right)$ looks a bit like a funnel with vertical symmetry, so it’s clear that if $x=4$ is a solution, $x=-4$ has to be, too.

-- Uncle Colin

* Edited 2015-09-26 to fix HTML.