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?       
 A: 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


*

*$\log(AB)=\log A+\log B$

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

*$\log\left(A^n\right)=n\log A$

*$\log(1)=0$
Notice that from (2) and (4) you get that $\log\left(\dfrac{1}{B}\right)=\log 1-\log B=-\log B$
A: Hint:
Product rule for Logarithms says that:
$$\log\prod_{k=1}^{n}a_k=\sum_{k=1}^{n}\log a_k$$
In the equation stated in your question:
$$\begin{aligned}\log F&=\log G+\log m+\log M+\log\dfrac{1}{r^2}\\ \log F &= \log \dfrac{GMm}{r^2}\\ \exp\log F&=\exp\log\dfrac{GMm}{r^2}\\  F&=G\cdot\dfrac{Mm}{r^2}\end{aligned}$$
A: The SINGLE most important rule of logarithms is:
$\log N + \log M = \log N\times M$.
This is because if $a = \log N; b=\log M$ then $10^a = N; 10^b = M$ so $10^{a+b} = 10^a\times 10^b = N\times M$ and so, by definition, $a+b = \log N\times M$.
From this simple rule we get $n \log M = \log (M^n)$ and $\log M -\log N = \log \frac MN$ and so on.
So...... 
Well, the basic rules of combining logarithms:  $\log a + \log b = \log ab$ will give us:
$\log{F} = \log{G} + \log{m} - \log\frac{1}{M} - 2\log{r} = \log \frac {Gm}{\frac 1Mr^2}=\log \frac {GMm}{r^2}$.
The log function is one to one so we know that for positive real numbers that $ a = b \iff \log a = \log b$.  
(If we need to convince ourselves of this: $$m= \log a = \log b = n\implies 10^m =10^{\log a} = a; 10^n = 10^{\log b}= b; 10^m = 10^n\implies a=b \implies \log a=\log b$$.)
So from here we get:
$F = \frac {GMm}{r^2}$.
