{{ item.displayTitle }}

No history yet!

Student

Teacher

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} A **logarithm** is the inverse of an exponential function. The logarithm of a positive number $m$ can be written as follows. The expression $g_{b}(m)$ is read as $g$ base $b$ of $m$

and states that raising $b$ to the $n_{th}$ power yields $m.$

$g_{b}(m)=n⇔b_{n}=m$

Here, $b$ is both the base of the logarithm and the base of the exponent. For example, the value of $n$ in $g_{4}(16)=n$ is given by the exponent to which $4$ would would be raised to result in $16.$
$g_{4}(16)=ng_{4}(16)=2 ⇔4_{n}=16⇔4_{2}=16 $
The *logarithmic form* and *exponential form* are equivalent.

A common logarithm is a logarithm of base $10.$ For example, $g_{10}(1000)$ is equal to $3$ because $10_{3}$ is equal to $1000.$

Since $g_{10}$ is used so often, it is sometimes written without a base. For positive values of $m,$ the common $g$ of $m$ can be defined as follows.

$g(m)=n⇔10_{n}=m$

For the following expressions, rewrite them in either logarithmic or exponential form. $g(100)=28_{2}=64g_{2}(32)=5e_{0}=1$

Show Solution

We will rewrite each expression one at a time using the relationship $g_{b}(m)=n⇔b_{n}=m.$ Notice that the first and third are written as logarithms and the second and fourth are written as exponents. Since a base is not written on the first logarithm, we know it's base $10.$ Furthermore, $2$ is the exponent. Thus, $g(100)=2⇔10_{2}=100.$ The same reasoning applies for the third expression, $g_{2}(32)=5.$ $g_{2}(32)=5⇔2_{5}=32$ Using the same relationship in the opposite way, we can rewrite the second and fourth expression. In $8_{2}=64,$ it can be seen that $8$ is the base and $2$ is the exponent. This means that $64$ is the value of which we take the logarithm. $8_{2}=64⇔g_{8}(64)=2.$ For the last expression, this yields $e_{0}=1⇔g_{e}(1)=0⇔ln(1)=0.$ To summarize, the following expressions are equivalent. $g(100)=28_{2}=64g_{2}(32)=5e_{0}=1 ⇔10_{2}=100⇔g_{8}(64)=2⇔2_{5}=32⇔ln(1)=0 $

Evaluate the following logarithms. $g_{3}(81)ln(e)g(10001 )$

Show Solution

When evaluating logarithms, it can be helpful to think about what the expression means. $g_{3}(81)$ asks the exponent to which $3$ must be raised to equal $81.$ Since, $3⋅3⋅3⋅3=81,$ $g_{3}(81)=4.$ The second expression, $ln(e),$ is a logarithm of base $e.$ It asks the exponent to which $e$ must be raised to equal $e.$ Since $e_{1}=e,$ $ln(e)=1.$ Notice that, unlike the other expressions, the last, which is of base $10,$ contains a fraction. Thus, we must consider the exponent to which $10$ is raised to equal $10001 .$ As it turns out, negative exponents yield fractions. $x_{-n}=x_{n}1 .$ It can be helpful to rewrite $1000$ as a power of $10.$ Since $1000=10_{3},$ $10001 =10_{3}1 =10_{-3}.$ Thus, $g(10001 )=-3.$

{{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

{{ exercise.headTitle }}

{{ 'ml-heading-exercise' | message }} {{ focusmode.exercise.exerciseName }}