algebra precalculus - How do I get from log F = log G + log m - log(1/M) - 2 log r to a solution withoug logs?

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?

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

1. $\log(AB)=\log A+\log B$


2. $\log\left(\dfrac{A}{B}\right)=\log A-\log B$


3. $\log\left(A^n\right)=n\log A$


4. $\log(1)=0$

Notice that from (2) and (4) you get that $\log\left(\dfrac{1}{B}\right)=\log 1-\log B=-\log B$