Logarithmic functions
If we find the inverse of an exponential function, we get another function, which we call a logarithmic function.
Consider an exponential equation,
$y=a^x$
Suppose we want to solve this equation for $x$. To make it easier, we write the solution as a new function, called the logarithmic function. It is written as
$x=\log_{a}y$
Here $a$ is called the base of the logarithm, which is the same as the base of the exponent.
Here we have two equations: the first is an exponential equation, and the second is the logarithmic form of the same equation. So, we can go from the exponential form to the logarithmic form and vice versa.
Example: Rewrite as a logarithmic equation, $3^4=81$
Solution:
First write the exponent; then put an equal sign and, on the other side, write the logarithm with its base followed by the number:
$4=\log_{3}81$
or
$\log_{3}81=4$
Evaluating logarithmic expressions
We can evaluate a logarithmic expression by forming a logarithmic equation and writing it as an exponential equation.
Example 1: Evaluate the logarithmic expression $\log_{8}64$.
Solution:
Take the expression as $x$,
$x=\log_{8}64$
Write this equation as an exponential equation,
$64=8^x$
We can write $64$ as $8^2$, therefore,
$8^2=8^x$
The bases on both sides are equal, so the powers must be equal. Therefore,
$2=x$
or
$x=2$
Since $x$ is the given logarithmic expression, we have
$\boxed{\log_{8}64=2}$
Example 2: Evaluate the logarithmic expression $\log_{4}\left(\dfrac{1}{64}\right)$.
Solution:
Take the expression as $x$,
$x=\log_{4}\left(\dfrac{1}{64}\right)$
Write this equation as an exponential equation,
$\dfrac{1}{64}=4^x$
We can write $\dfrac{1}{64}$ as $4^{-3}$ (since $64=4^3$), therefore,
$4^{-3}=4^x$
The bases on both sides are equal, so the powers must be equal. Therefore,
$x=-3$
Since $x$ is the given logarithmic expression, we have
$\boxed{\log_{4}\left(\dfrac{1}{64}\right)=-3}$
Properties of logarithm
$\log_{a}1=0$
$\log_{a}a=1$
$\log_{a}a^p=p$
$a^{\log_{a}x}=x$
Product property of logarithm
$\log_{a}{XY}=\log_{a}{X}+\log_{a}{Y}$
Quotient property of logarithm
$\log_{a}\left(\dfrac{X}{Y}\right)=\log_{a}{X}-\log_{a}{Y}$
Power property of logarithm
$\log_{a}x^p=p\log_{a}x$
Example 1: Express the following as a single logarithm,
a. $\log_{7}5+\log_{7}11$
b. $\log_{6}13-\log_{6}4$
Solution:
a. $\log_{7}5+\log_{7}11=\log_{7}(5\cdot 11)$
$\hphantom{a. \log_{7}5+\log_{7}11}=\log_{7}{55}$.
b. $\log_{6}13-\log_{6}4=\log_{6}\left(\dfrac{13}{4}\right)$.
Example 2: Expand the following logarithm using the properties of logarithms,
$\log \left(x^{10}y^3z\right)$.
Solution:
Use the product rule,
$\log \left(x^{10}y^3z\right)= \log x^{10}+\log y^3+\log z$.
Apply the power rule,
$\hphantom{\log \left(x^{10}y^3z\right)}=10\log x+3\log y+\log z$.
Example 3: Expand the following logarithm using the properties of logarithms,
$\log \left(\dfrac{x^{4}y^7}{z^6}\right)$.
Solution:
Use the product and the quotient rule,
$\log \left(\dfrac{x^{4}y^7}{z^6}\right)=\log x^{4}+\log y^7-\log z^6$.
Apply the power rule,
$\hphantom{\log \left(\dfrac{x^{4}y^7}{z^6}\right)}=4\log x+7\log y-6\log z$.
Example 4: Write as a single logarithm,
$5\log_{b}x-4\left(\log_{b}u-3\log_{b}y\right)$
Solution:
Remove the parentheses,
$5\log_{b}x-4\left(\log_{b}u-3\log_{b}y\right)$
$\hphantom{00000}=5\log_{b}x-4\log_{b}u+12\log_{b}y$
Apply the power rule,
$\hphantom{00000}=\log_{b}x^5-\log_{b}u^4+\log_{b}y^{12}$
Apply the product and the quotient rule to combine,
$\hphantom{00000}=\log_{b}\left(\dfrac{x^5y^{12}}{u^4}\right)$
Compound interest problems
Suppose you invest an amount $P$ at an annual interest rate $r$. Then the amount in your account after a certain time period is given by
$A(t)=P\left(1+\dfrac{r}{n}\right)^{nt}$
where $t$ is the number of years and $n$ is the number of times per year the interest is compounded. For example, $n=1$ if the interest is compounded annually, and $n=12$ if it is compounded monthly.
The interest can also be compounded continuously. If the interest is compounded continuously, then
$A(t)=Pe^{rt}$
Example: John deposited $\$3900$ in a bank account with an interest rate of $1.8\%$, compounded each year. How much money will be in his account after $6$ years? Assume there are no withdrawals. Round your answer to the nearest cent.
Solution:
The original amount deposited is $P=3900$.
Number of years, $t=6$
Interest rate is $r=1.8\%=1.8/100=0.018$
The interest is compounded each year, that is, annually; therefore, $n=1$.
The formula to find the amount at the end of the period is
$A(t)=P\left(1+\dfrac{r}{n}\right)^{nt}$
Substituting the values,
$A(t)=3900\left(1+\dfrac{0.018}{1}\right)^{1\cdot 6}$
$\hphantom{A(t)}=3900(1.018)^6$
$\hphantom{A(t)}=4340.62$
$\hphantom{A(t)}\boxed{Answer=\$ 4340.62}$
Changing the base of a logarithm
A logarithm with base $10$ is called the common logarithm. We can find the value of a common logarithm or a natural logarithm using a calculator. But if the base is not $10$ or $e$, we need to change it to one of these so that we can use the calculator.
If you have a logarithm with base $b$ and want to change it to base $a$, divide the base-$a$ logarithm of $x$ by the base-$a$ logarithm of $b$:
$\log_{b}x=\dfrac{\log_{a}x}{\log_{a}b}$
Example 1: Use the change-of-base formula to evaluate,
$\log_{4}78$
Round your answer to the nearest thousandth.
Solution:
Change the base from $4$ to $10$,
$\log_{4}78=\dfrac{\log_{10}78}{\log_{10}4}$
Use a calculator,
$\hphantom{\log_{4}78}=\dfrac{1.8921}{0.6021}$
$\hphantom{\log_{4}78}=3.143$
Example 2: Use the change-of-base formula to evaluate,
$\log_{2/7}\left(\dfrac{1}{11}\right)$
Round your answer to the nearest thousandth.
Solution:
Change the base to $10$,
$\log_{2/7}\left(\dfrac{1}{11}\right)=\dfrac{\log \left(\dfrac{1}{11}\right)}{\log \left(\dfrac{2}{7}\right)}$
Use a calculator,
$\hphantom{\log_{2/7}\left(\dfrac{1}{11}\right)}=\dfrac{-1.0414}{-0.5441}$
$\hphantom{\log_{2/7}\left(\dfrac{1}{11}\right)}=1.914$