# 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?

• Since each $\log$ is in the same base (ten) you can cancel them. That is, if $\log_b(x)=\log_b(y)$ then $x=y$. P.S. The converse is also true since $\log$ is an inyective function. Mar 19, 2019 at 17:17

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

• Nice clear answer. You may like to point out that $a = b \iff \log a = \log b$ so that $\log F = \log \frac {GMm}{r^2}$ would also mean $F = \frac {GMm}{r^2}$. Mar 19, 2019 at 18:27
• @fleablood Good point. Mar 19, 2019 at 20:04

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}

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:

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