7.08 Properties of logarithms

Laws of exponents and laws of logarithms

Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.

$\log_b1=0$logb​1=0

$\log_bb=1$logb​b=1

Why are the above true? To find out, change the logarithm to an exponential: $\log_b1=0$logb​1=0 becomes $b^0=1$b0=1  and $\log_bb=1$logb​b=1 becomes $b^b=1$bb=1. 

Now, lets review the inverse property:

$\log_bb^x=x$logb​bx=x

$b^{\log_bx}=x$blogb​x=x, $x>0$x>0

Why is this true? Convert each into its opposite: $\log_bb^x=x$logb​bx=x becomes $b^x=b^x$bx=bx and $b^{\log_bx}=x$blogb​x=x becomes $\log_bx=\log_bx$logb​x=logb​x. 

Worked example

Example 1

What is the value of $x$x for the equation $\log_33x=\log_3\left(2x+5\right)$log3​3x=log3​(2x+5)?

Think: Do the logarithms have the same base? Remember, we can not work with logarithms with unlike bases. They do have the same base. So, that means, just as with exponential equations with the same bases, we can set the arguments equal to each other.

Do:

$\log_33x$log3​3x	$=$=	$\log_3\left(2x+5\right)$log3​(2x+5)	(Given)
$3x$3x	$=$=	$2x+5$2x+5	(Use one-to-one property to set arguments equal)
$x$x	$=$=	$5$5	(Subtract both sides by $2x$2x)

 

Product rule for logarithms

Recall the product rule for exponents:

$a^ma^n=a^{m+n}$aman=am+n

The product rule for logarithms is similar:

$\log_bmn=\log_bm+\log_bn$logb​mn=logb​m+logb​n

Why is this true? Let's use the rules we already know to prove this:

$\log_bmn$logb​mn	$=$=	$\log_bm+\log_bn$logb​m+logb​n	(Given)
$b^{\log_bm+\log_bn}$blogb​m+logb​n	$=$=	$mn$mn	(Rewrite as an exponential)
$b^{\log_bm}b^{\log_bn}$blogb​mblogb​n	$=$=	$mn$mn	(Use laws of exponent: product rule)
$mn$mn	$=$=	$mn$mn	(Use inverse property for logarithms)

 

The product rule for logarithms can be applied repeatedly. The expression $\log_btuv$logb​tuv can be rewritten as $\log_bt+\log_bu+\log_bv$logb​t+logb​u+logb​v.

Worked example

example 2

Rewrite $\log_56x$log5​6x as two logarithms.

Think: Since there is a product in the logarithm, we can use the product rule for logarithms.

Do: So, using the product rule for logarithms, we can rewrite $\log_56x$log5​6x in the form:

$\log_56+\log_5x$log5​6+log5​x

Quotient rule for logarithms

Recall the quotient rule for exponents:

$$

The quotient rule for logarithms is similar:

$\log_b\left(\frac{m}{n}\right)=\log_bm-\log_bn$logb​(mn​)=logb​m−logb​n

Why is this true? Lets take a look using properties we already know:

$\log_b\left(\frac{m}{n}\right)$logb​(mn​)	$=$=	$\log_bM-\log_bN$logb​M−logb​N	(Given)
$b^{\log_bm-\log_bn}$blogb​m−logb​n	$=$=	$\frac{m}{n}$mn​	(Rewrite as an exponential)
$\frac{b^{\log_bm}}{b^{\log_bn}}$blogb​mblogb​n​	$=$=	$\frac{m}{n}$mn​	(Use laws of exponent: quotient rule)
$\frac{m}{n}$mn​	$=$=	$\frac{m}{n}$mn​	(Use inverse property for logarithms)

 

Worked example 

example 3

Rewrite the logarithmic expression $\log_310-\log_32$log3​10−log3​2 as one logarithmic expression.

Think: Since the two logarithms have the same base and one is being subtracted from the other, we can use the quotient rule for logarithms.

Do: To use the quotient rule for logarithms, we can divide the arguments:

$\log_310-\log_32$log3​10−log3​2	(Given)
$\log_3\left(\frac{10}{2}\right)$log3​(102​)	(Use quotient rule for logarithms)
$\log_35$log3​5	(Simplify the argument)

Practice questions

Question 1

Question 2

Question 3

 

Power rule for logarithms

We have already seen how to simplify logarithms using the product and quotient properties. Through the definition of logarithms, recall that $x=a^m$x=am and $m=\log_ax$m=loga​x are equivalent. This definition can be used to discover some more helpful properties of logarithms such as the power property.

 

Exploration

Let's simplify $\log_a\left(x^2\right)$loga​(x2) using the logarithmic properties that we already know.

$\log_a\left(x^2\right)$loga​(x2)	(Given)
$\log_a\left(x\times x\right)$loga​(x×x)	(Rewrite $x^2$x2as a product, $x\times x$x×x)
$\log_ax+\log_ax$loga​x+loga​x	(Use the product rule for logarithms, $\log_a\left(xy\right)=\log_ax+\log_ay$loga​(xy)=loga​x+loga​y
$2\log_ax$2loga​x	(Collect logarithms with the same base and variables.)

 

Logarithms with powers can also be simplified using the power rule for logarithms. This property can be used for any values of the power $n$n.

Worked example

example 4

Rewrite $\log_2\left(x^b\right)$log2​(xb) using properties of logarithms. Write your answer without any powers.

Think: The subject of the logarithm has a power, this means we can use the power rule for logarithms,$\log_a\left(x^n\right)=n\log_ax$loga​(xn)=nloga​x.

What do the values of $a$a and $n$n represent?

Do: In this case $a=2$a=2 and $n=b$n=b. So we can bring the power down to the front, and then multiply it with the logarithm.

$\log_2\left(x^b\right)$log2​(xb)

$=$=

$b\log_2x$blog2​x

 

Practice questions

Question 4

Question 5 

Question 6

Condensing and expanding logarithmic expressions

When viewed together, the product rule, quotient rule and power rule are often called "laws of logs." More than one rule can be applied to simplify an expression.

Worked examples

example 5

Simplify the expression $\log_3\left(100x^3\right)-\log_3\left(4x\right)$log3​(100x3)−log3​(4x), writing your answer as a single logarithm.

Think: Each logarithm in the expression has the same base, so we can express the difference as a single logarithm using the quotient rule.

Do: To use the quotient rule, we divide the two arguments as follows:

$\log_3\left(100x^3\right)-\log_3\left(4x\right)$log3​(100x3)−log3​(4x)	(Given)
$\log_3\left(\frac{100x^3}{4x}\right)$log3​(100x34x​)	(Use the quotient rule)
$\log_3\left(25x^2\right)$log3​(25x2)	(Simplifying the argument)
$\log_3\left(\left(5x\right)^2\right)$log3​((5x)2)	(Rewriting the argument as a power)
$2\log_3\left(5x\right)$2log3​(5x)	(Using the power rule)

 

example 6

Expand the expression: $\ln\frac{x^2y^3}{z^4}$lnx2y3z4​

Think: List the operations within the argument: exponents, multiplication, division. We are going to start expanding with division because of the sensitivity to order. 

Do:

$\ln\frac{x^2y^3}{z^4}$lnx2y3z4​	(Given)
$\ln x^2y^3-\ln z^4$lnx2y3−lnz4	(Use the quotient rule)
$\ln x^2+\ln y^3-\ln z^4$lnx2+lny3−lnz4	(Use the product rule)
$2\ln x+3\ln y-4\ln z$2lnx+3lny−4lnz	(Use power rule)

Practice questions

Question 7

Question 8

 

Evaluate logarithmic functions

Logarithmic functions can be re-written in exponential form in order to evaluate the function more easily.

Remember that, since an exponent can never evaluate to a negative number, the value inside a logarithm can never be negative. This means that logarithmic functions can not be evaluated for negative values. The domain of logarithmic functions will be discussed further when we look at their graphs.

Worked examples

example 7

Evaluate the following function when $x=0.01$x=0.01: $f\left(x\right)=\log x$f(x)=logx

Think: If $f\left(x\right)=\log x$f(x)=logx, then, assuming base $10$10. Also, rewrite the decimal as an exponent with base of 10.

Do:  

$f\left(0.01\right)$f(0.01)	$=$=	$\log\left(0.01\right)$log(0.01)
	$=$=	$\log\left(10^{-2}\right)$log(10−2)
	$=$=	$-2$−2

Example 8

Evaluate $f\left(x\right)=\log_2\left(3x-2\right)$f(x)=log2​(3x−2) when $x=22$x=22

Think: If $f\left(x\right)=\log_2\left(3x-2\right)$f(x)=log2​(3x−2), then the base is $2$2. Substitute the value of $x$x with $22$22, then rewrite value as an exponent with a base of $2$2.

Do: 

$f\left(22\right)$f(22)	$=$=	$\log_2\left(3\times22-2\right)$log2​(3×22−2)
	$=$=	$\log_2\left(64\right)$log2​(64)

Rewrite as an exponent with base of 2 ($2^6=64$26=64), therefore $x=6$x=6.

example 9

Evaluate $f\left(x\right)=\log_7\left(2-x\right)$f(x)=log7​(2−x) when $x=-5$x=−5.

Think: If $f\left(x\right)=\log_7\left(2-x\right)$f(x)=log7​(2−x), then the base is $7$7. Substitute the value of $x$x with $-5$−5, then rewrite value as an exponent with a base of $7$7.

Do:

$f\left(-5\right)$f(−5)	$=$=	$\log_7\left(2-(-5)\right)$log7​(2−(−5))
	$=$=	$\log_7\left(7^1\right)$log7​(71)
	$=$=	$1$1

Reflect: Note here that the argument $\left(2-x\right)$(2−x) must be kept positive. Thus the domain of this function is found by setting $\left(2-x\right)>0$(2−x)>0, from which we can conclude that $x<2$x<2. 

Practice questions

question 9

Question 10