11.07 Logarithm laws

In the same way that there are index laws which allow us to simplify exponential expressions, there are logarithm laws that allow us to simplify logarithmic expressions. In fact, each logarithm law is a consequence of an index law.

First, consider the definition that if $y=B^x$y=Bx then $x=\log_By$x=logB​y. It follows that,

$\log_BB^x=x$logB​Bx=x

Substituting $x=0$x=0 and $x=1$x=1 gives the following special cases:

$\log_B1=0$logB​1=0

$\log_BB=1$logB​B=1

For the following proofs, we will let $p=B^m$p=Bm and $q=B^n$q=Bn so that $\log_Bp=m$logB​p=m and $\log_Bq=n$logB​q=n. Note that for any $B\ne0$B≠0 there will be some values of $m,n,p,$m,n,p, and $q$q which makes these equations true.

$B^m\times B^n$Bm×Bn	$=$=	$B^{m+n}$Bm+n	From the index law
$\log_BB^m\times B^n$logB​Bm×Bn	$=$=	$\log_BB^{m+n}$logB​Bm+n	Taking the logarithm base $B$B from both sides
$\log_BB^m\times B^n$logB​Bm×Bn	$=$=	$m+n$m+n	From the definition of logarithms
$\log_Bpq$logB​pq	$=$=	$\log_Bp+\log_Bq$logB​p+logB​q	Substituting values of $m,n,p,$m,n,p, and $q$q

 

$\frac{B^m}{B^n}$BmBn​	$=$=	$B^{m-n}$Bm−n	From the index law
$\log_B\frac{B^m}{B^n}$logB​BmBn​	$=$=	$\log_BB^{m-n}$logB​Bm−n	Taking the logarithm base $B$B from both sides
$\log_B\frac{B^m}{B^n}$logB​BmBn​	$=$=	$m-n$m−n	From the definition of logarithms
$\log_B\frac{p}{q}$logB​pq​	$=$=	$\log_Bp-\log_Bq$logB​p−logB​q	Substituting values of $m,n,p,$m,n,p, and $q$q

 

$\left(B^m\right)^n$(Bm)n	$=$=	$B^{mn}$Bmn	From the index law
$\log_B\left(B^m\right)^n$logB​(Bm)n	$=$=	$\log_BB^{mn}$logB​Bmn	Taking the logarithm base $B$B from both sides
$\log_B\left(B^m\right)^n$logB​(Bm)n	$=$=	$mn$mn	From the definition of logarithms
$\log_Bp^n$logB​pn	$=$=	$n\log_Bp$nlogB​p	Substituting values of $m$m and $p$p