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

Question

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 1

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 $\log\left(\dfrac{1}{B}\right)=\log 1-\log B=-\log B$

Answer 2

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: $$\begin{aligned}\log F&=\log G+\log m+\log M+\log\dfrac{1}{r^2}\\ \log F &= \log \dfrac{GMm}{r^2}\\ \exp\log F&=\exp\log\dfrac{GMm}{r^2}\\ F&=G\cdot\dfrac{Mm}{r^2}\end{aligned}$$

Answer 3

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:

$F = \frac {GMm}{r^2}$.