When can using a logarithm make solving an equation easier?

[bolzano95]

Homework Statement

Well, there is a physics problem I was solving and it is really interesting how it is officially solved.

We take a small weight and hang it on a steel wire. For how much does the oscillation time change if the temperature of this wire raises for 10K?

I looked up solution and it is solved like this:

Homework Equations

The weight on the steel wire is like a mathematical pendulum.
Therefore $t_0 = 2\pi \sqrt{\frac{L}{g}}$. Now logarithm and differentiate:
$ln {t_0} = ln {2\pi} + 1/2 ln {L} -1/2 ln{g}$ and after differentiation
$\frac{dt_0}{t_0}= \frac{1}{2} \frac{dL}{L}$.

Because of the temperature change $dt$ the steel wire is longer for $dL =\alpha L dT$.
The relative change of oscillation time is then $\frac{dt_0}{t_0}= \frac{1}{2} \alpha dT$.

The Attempt at a Solution

In my solution process there was nothing of logarithms or differentiation. Of course my result was also false. But I am not interested in where I did a mistake, there is more important question her:

What is this special new solving approach? It is the first time I see solving it and it frustrates me, because I don't understand the logic behind it.
When can I use it in the future? Also from where does it come?
P.S Even in my wildest dreams I would not use this solving method.

 

[kuruman]

It's only a quick way to get the fractional change of variable $x$ which is $\frac{dx}{x}$. You could have obtained the same result by plain differentiation.

For example in this case you would do $$dt_0=\frac{2\pi}{\sqrt{g}}\frac{1}{2}\frac{1}{\sqrt{L}}dL$$Then divide by $t_0$ or multiply by $t_0^{-1} ~$ to get$$\frac{dt_0}{t_0}=\frac{2\pi}{\sqrt{g}}\frac{1}{2}\frac{1}{\sqrt{L}}dL\times \frac{\sqrt{g}}{2\pi\sqrt{L}}=\frac{1}{2}\frac{dL}{L}.$$

For whatever it's worth, this is the first time I see the logarithmic method for doing this. If it's confusing you, don't use it.

 

[quinoa19]

There is a technique called Logarithmic differentiation, which is applied here. It sometimes leads to answer faster as in this case. Of course the answer should be same regardless of whether it is used or not.