I've been self-studying from Stroud & Booth's excellent "Engineering Mathematics", and am currently on the "Algebra" section. I understand everything pretty well, except when it comes to the problems then I am asked to express an equation that uses logs, but without logs, as in:
$$\log{F} = \log{G} + \log{m} – \log\frac{1}{M} – 2\log{r}$$
They don't cover the mechanics of doing things like these very well, and only have an example or two, which I "kinda-sorta" barely understood.
Can anyone point me in the right direction with this and explain how these are solved?
Best Answer
Using some rules of logarithms you get $\quad-\log\dfrac{1}{M}=+\log M$ and $-2\log r=-\log r^2=+\log \dfrac{1}{r^2}$
So you have
\begin{eqnarray} \log{F} &=& \log{G} + \log{m} + \log M + \log{\dfrac{1}{r^2}}\\ \log{F} &=& \log{\left(GmM\cdot\dfrac{1}{r^2}\right)}\\ \log{F} &=& \log{\frac{mMG}{r^2}}\\ F &=&\frac{mMG}{r^2} \end{eqnarray}
The last step hinges upon the fact that logarithm functions are one-to-one functions. If a function $f$ is one-to-one, then $f(a)=f(b)$ if and only if $a=b$. Since $\log$ is a one-to-one function, it follows that $\log A=\log B$ if and only if $A=B$.
ADDENDUM: Here are a few rules of logarithms which you may need to review
Notice that from (2) and (4) you get that $\log\left(\dfrac{1}{B}\right)=\log 1-\log B=-\log B$