Evaluate logarithmic expressions

 

Simplifying using the definition

Recall that the log expression $\log_ba$logb​a can be thought of as the number that the base $b$b must be raised to in order to become $a$a. This is a key understanding of the expression.

If, for example, $10$10 is raised to the power $3$3 to become $1000$1000, then $3$3 is said to be the base $10$10 logarithm of $1000$1000. It is that simple - the $\log_{10}1000=3$log10​1000=3. 

There are working rules that enable us to simplify expressions involving logs, however it is often best to consider the above definition first. 

Consider simplifying separately the three expressions given as 

$\log_381$log3​81

$\log_5\sqrt{5}$log5​√5 and 

$\log_51$log5​1.

 

The first simplifies to $4$4 because $3^4=81$34=81.

The second simplifies to $\frac{1}{2}$12​ because $5^{\frac{1}{2}}=\sqrt{5}$512​=√5.

The last simplifies to $0$0 because all non-zero numbers raised to the power $0$0 become $1$1. 

 

What about an expression like $\log_{0.5}2$log0.5​2? We ask what must $\frac{1}{2}$12​ be raised to, to become $2$2? Algebraically, we need to find $x$x such that  $\frac{1}{2}^x=2$12​x=2. We know that $\frac{1}{2}^x=\left(2^{-1}\right)^x=2^{-x}$12​x=(2−1)x=2−x so this tells us that $x=-1$x=−1. Thus $\log_{\frac{1}{2}}\left(2\right)=-1$log12​​(2)=−1. 

Expressions like $\log_3\left(4\right)$log3​(4) are evaluated using a calculator. The change of base rule, if required, can alter the expression to a quotient of base $10$10 logarithms so that $\log_3\left(4\right)=\frac{\log_{10}\left(4\right)}{\log_{10}\left(3\right)}=1.26186$log3​(4)=log10​(4)log10​(3)​=1.26186 to $5$5 decimal places.

Simplifying using the working rules

The working rules can be listed as follows: 

More detail on the log laws can be found here. 

Each of these can be proven using the definition.

For example we can show that the first rule is true by setting $x=\log_bm$x=logb​m and $y=\log_bn$y=logb​n. Then equivalently $m=b^x$m=bx and $n=b^y$n=by so that $mn=b^{x+y}$mn=bx+y and this means that $x+y=\log_bmn$x+y=logb​mn. Thus $\log_bmn=\log_bm+\log_bn$logb​mn=logb​m+logb​n.  Similar strategies are used for other rules.

So for example the expression  $\log_381$log3​81 simplified by the definition above can be simplified using the working rules. Thus, $\log_381$log3​81 becomes $\log_3\left(3\right)^4$log3​(3)4 which by working rule $3$3 becomes $4\log_33$4log3​3 which is simply $4$4.

An expression like $\log_2\sqrt{256}-\log_2\sqrt{128}$log2​√256−log2​√128 can be simplified as follows:

$\log_2\sqrt{256}-\log_2\sqrt{128}$log2​√256−log2​√128	$=$=	$\log_2\sqrt{\frac{256}{128}}$log2​√256128​
	$=$=	$\log_2\sqrt{2}$log2​√2
	$=$=	$\log_2\left(2^{\frac{1}{2}}\right)$log2​(212​)
	$=$=	$\frac{1}{2}$12​

We can also evaluate certain logarithm expressions knowing other expressions. For example, if we know that $\log_b2=0.3010$logb​2=0.3010 and $\log_b3=0.4771$logb​3=0.4771, then we also know that $\log_b72$logb​72, which can be expressed as $\log_b\left(2^3\times3^2\right)$logb​(23×32), becomes $3\log_b2+2\log_b\left(3\right)=3\left(0.3010\right)+2\left(0.4771\right)$3logb​2+2logb​(3)=3(0.3010)+2(0.4771) or $1.8572$1.8572.

Worked Examples

QUESTION 1

QUESTION 2