Chapter 3.5, Problem 69E

Solution

To prove: Prove that, if $\frac{u}{v}={10}^n$ for $\text{u}>0$ and $\text{v}>0$ , then $\log u–\log v=n$ and hence explain how this result relates to powers of ten and orders of magnitude.

Given information:

  $\frac{u}{v}={10}^n$ for $\text{u}>0$ and $\text{v}>0$ .

Formula used:

  $\begin{array}{l}\log \left(\frac{a}{b}\right)=\log \left(a\right)-\log \left(b\right)\\ \log \left(a^n\right)=n\log \left(a\right)\end{array}$

Proof:

Let $u,v$  represent positive real numbers. Let $n$  represent a real number such that;

  $\frac{u}{v}={10}^n$

Take the common logarithm of both sides:

  $\log \left(\frac{u}{v}\right)=\log \left({10}^n\right)$

Apply the Quotient Rule to the left-hand side and the Power Rule to the right-hand side:

  $\log \left(u\right)-\log \left(v\right)=n\log \left(10\right)$

As $\log \left(10\right)=1$ :

  $\log u–\log v=n$

This concludes the proof.

When talk about order of magnitude, usually referring to the power of ten of the given value. Still using the variables from above, let  $a$ and  $b$ represent real numbers such that,

  $\begin{array}{l}a\text{}=\log u\\ b=\log v\text{}\end{array}$

By the definition of a logarithm,

  $\begin{array}{l}u={10}^a\\ v={10}^b\text{}\end{array}$

And then the difference between $a$ and  $b$ is what we would refer to as the difference in orders of magnitude.