# Speak Math Now!/Week 9: Six rules of Exponents/Logarithms

Abstract

This is a subpage to Speak Math Now!/Week 9: Six rules of Exponents, which lays out the properties of exponents in a painstaking fashion that is appropriate for students learning the subject for the first time. Here, we show how five of the six rules of exponents can be modified to "derive" the analogous rules for logarithms. We begin by defining the logarithm as the inverse of the exponential function, using the reflection of the a function through the line x=y to establish the logarithm's graph for base b>1. Then we show how five of the six rules in the parent page yield identities useful for manipulating logarithms. Finally, we derive the formula for changing a logarithm's base. A comparison with two derivations already on Wikipedia suggests that this derivation might be easier to remember.

## The logarithm is the exponent Displays symmetry between log and exponential functions. If you study this diagram long enough, you will understand why the inverse of any function is obtained by reflecting a graph of the function through line S. See also w:Inverse_function.

While teaching college physics I often encountered students who had forgotten the defintion of the logarithm. Then, after a few years of teaching only conceptual courses in physical science and astronomy, it almost happened to me. Then, I read in one of those textbooks that allows me to reconstruct almost everything I had leaened about logarithms: The logarithm is the exponent. If $n$ is the exponent and b is the base, then we have:

$z=b^{n}\Leftrightarrow n=\log _{b}(z)$ (A)

The $\Leftrightarrow$ (if and only if) symbol requires two restrictions on the variables. Great complications arise unless we insist that $(n,z,b)$ are all real numbers. Furthermore:

• The base $b$ can be any positive real number not equal to one: $0 and $b\neq 1$ .
• We only take the logarithm of positive numbers: $z>0.$ Two important facts about logarithms emerge if set $n=0$ or $n=1$ . If $n=0$ , then $x=1,$ which implies that:

$\log _{b}1=0$ (B)

If $n=1$ , then $z=b,$ which leads to:

$\log _{b}b=1$ (C)

### The six rules revisited

The parent page introduces six rules the are useful when dealing with exponents. Here we show that five of them yield useful rules for logarithms.

#Rule 1 (Product of Powers) tells us that $b^{m}b^{n}=b^{m+n}.$ To convert this into an identity about logarithms, we set $u=b^{m},$ $v=b^{n},$ and $w=b^{n+m}.$ On a board or piece of paper I would write this as:

$\underbrace {b^{m}} _{u}\underbrace {b^{n}} _{v}=\underbrace {b^{m+n}} _{w=uv}$ Now apply the rule introduced after (A) to all three exponents in the above equation:

${\begin{array}{lcl}m&=&\log _{b}(u)\\n&=&\log _{b}(v)\\m+n&=&\log _{b}(w)=\log _{b}(uv)\\\end{array}}$ .

We conclude that rule 1 for exponents yields this rule for logarithms:

$\log _{b}u+\log _{b}v=\log _{b}(uv)$ (1)

At some level, the rigorous derivation leading to Equation (1) is essential. But for many of us, it is more useful to be able to recover these logarithmic identities using a less rigorous approach. Instead of deriving a result, we use simple examples to "guess" the correct result. For example:

#Rule 2 (Power to a Power) tells us that $(b^{m})^{n}=b^{mn}.$ To recover the logarithemetic version of this identity, we set $u=v$ in (1) to conclude that $\log(u^{2})=2\log u.$ For small values of $n$ we can verify that:

$n\log _{b}(x^{n})=n\log _{b}x$ (2)

#Rule 3 (Multiple Power Rules) allows us to combine rules 2 and 3: $6(x^{3}y)(x^{2}y)^{3}$ $=6(x^{3}y)(x^{6}y^{3})$ $=6x^{9}y^{4}$ implies that:

$\log _{b}\left[6(x^{3}y)(x^{2}y)^{3}\right]=log_{b}6+9log_{b}x+4\log _{b}y$ (3)

#Rule 4 (Quotient of Powers) From $b^{m}/b^{n}=b^{m-n}$ we obtain:

$\log _{b}\left({\frac {m}{n}}\right)=log_{b}(m)-log_{b}(n)$ (4)

#Rule 5 (Power of a Quotient) It is not obvious what to do with $(a/b)^{m}=a^{m}/b^{m}.$ Let's try taking the logarithm to base $a/b$ of both sides. From equation (A) we can set the LHS equal to the RHS, where

${\text{LHS}}=\log _{a/b}\left[\left({\frac {a}{b}}\right)^{m}\right]=m\log _{a/b}\left({\frac {a}{b}}\right)=m$ ${\text{RHS}}=m\log _{a/b}a-m\log _{a/b}b$ ${\text{LHS}}={\text{RHS}}\Rightarrow 1=\log _{a/b}a-\log _{a/b}b=\log _{a/b}\left({\frac {a}{b}}\right)=1$ At this point it seems nothing results from this attempt to view rule 5 from the perspective of the logarithm.

${\text{There is perhaps no logarithmetic analog to rule 5}}$ (5)

#Rule 6 (Negative Exponents) From $b^{-n}=1/b^{n}$ we obtain a formula already obvious from (4) above:

$\log _{b}(1/n)=-\log _{b}(n)$ (6)

## Change of base

We begin with the most obvious statement involving exponents of two different bases:

$b^{n}=B^{N}=z$ Ignore the $z$ , randomly pick one of the bases, and solve for the other base:

$b=B^{N/n}$ Now we take a logarithm of each side. The beauty of this proof is that it does which base you choose for this step:

If $B$ is the base: $\log _{B}(b)=\log _{B}(B^{N/n})={\frac {N}{n}}\log _{B}(B)={\frac {N}{n}}={\frac {\log _{B}(x)}{\log _{b}(x)}},$ which leads to:

$\log _{b}(x)={\frac {\log _{B}(x)}{\log _{B}(b)}}$ If $b$ is the base: $\log _{b}(b)=1=\log _{b}(B^{N/n})={\frac {N}{n}}\log _{b}(B)={\frac {\log _{B}(x)}{\log _{b}(x)}}\log _{b}(B),$ which leads to:

$\log _{B}(x)={\frac {\log _{b}(x)}{\log _{b}(B)}}$ It doesn't matter which formula you derive because one can be quickly obtained from the other by exchanging the capital and lower case letters $(b\rightleftarrows B)$ . This symmetry also leads to an interesting fact about any pair of bases:

$\log _{b}(B)\cdot \log _{B}(b)=1$ ### Comparisons with other derivations on Wikipedia

I find the other derivations found on Wikipedia harder to remember:

Apply logk to both sides of $x=b^{\log _{b}x}.$ Note: This derivation is shorter, but requires that one remembers the identity $x=b^{\log _{b}x}.$ .
Consider the equation $b^{c}=a$ , and take logarithm base $d$ of both sides to obtain: $\log _{d}b^{c}=\log _{d}a.$ Simplify and solve for $c$ to obtain $c\log _{d}b=\log _{d}a$ and then $\ldots$ . Note: I find this derivation far too complicated.