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A logarithm is the inverse function to an exponential function. It is defined as the exponent to which a fixed base must be raised to yield a given number.
Conversion Between Logarithmic and Exponential Form
There is actually a simple way to solve difficult questions that involve logarithmic or exponential equations. The key is to remember that an exponential equation is just a special case of a logarithmic equation, so it’s possible to convert an exponential equation into a logarithmic equation. So how do we convert the equation?
${{a}^{b}}=c\underset{{}}{\longleftrightarrow}b={{\log }_{a}}c$
$a$: base
$b$: index
Important Notes:
- $a>0,c>0$
${{\log }_{10}}x=\lg x$
${{\log }_{e}}x=\ln x$
Practice Question 1
Conversion to Logarithmic Form
Convert each of the following to logarithmic form.
(a)
${{x}^{2y}}=4$
(a) ${{x}^{2y}}=4$
(b)
${{5}^{x}}=12$
(b) ${{5}^{x}}=12$
(c)
${{\left( 4x \right)}^{5-p}}=a$
(c) ${{\left( 4x \right)}^{5-p}}=a$
(a)
(b)
(c)
(a)
$\begin{aligned}
{{x}^{2y}}&=4 \\
& \Downarrow \\
2y&={{\log }_{x}}4
\end{aligned}$
(b)
$\begin{aligned}
{{5}^{x}}&=12 \\
& \Downarrow \\
x&={{\log }_{5}}12
\end{aligned}$
(c)
$\begin{aligned}
{{\left( 4x \right)}^{5-p}}&=a \\
& \Downarrow \\
5-p&={{\log }_{4x}}a
\end{aligned}$
(a)
(b)
(c)
(a)
$\begin{aligned}
{{x}^{2y}}&=4 \\
& \Downarrow \\
2y&={{\log }_{x}}4
\end{aligned}$
(b)
$\begin{aligned}
{{5}^{x}}&=12 \\
& \Downarrow \\
x&={{\log }_{5}}12
\end{aligned}$
(c)
$\begin{aligned}
{{\left( 4x \right)}^{5-p}}&=a \\
& \Downarrow \\
5-p&={{\log }_{4x}}a
\end{aligned}$
Practice Question 2
Conversion to Exponential Form
Solve each of the following equations.
(a)
$\ln 4x=5$
(a) $\ln 4x=5$
(b)
${{\log }_{x}}16=4$
(b) ${{\log }_{x}}16=4$
(c)
$\ln (3x)=6$
(c) $\ln (3x)=6$
(d)
${{\log }_{k}}81=2$
(d) ${{\log }_{k}}81=2$
(a)
(b)
(c)
(d)
(a)
$\begin{aligned}
\ln 4x&=5 \\
{{\log }_{\text{e}}}4x&=5 \\
4x&={{\text{e}}^{5}} \\
x&=\frac{1}{4}{{\text{e}}^{5}}
\end{aligned}$
(b)
$\begin{aligned}
{{\log }_{x}}16&=4 \\
{{x}^{4}}&=16 \\
x&=\pm \sqrt[4]{16} \\
x&=2\,\,\text{or}\,\,-2
\end{aligned}$
$x=-2$ (rej, as $x>0$)
$\therefore x=2$
(c)
$\begin{aligned}
\ln \left( 3x \right)&=6 \\
{{\log }_{\text{e}}}\left( 3x \right)&=6 \\
{{\text{e}}^{6}}&=3x \\
x&=\frac{1}{3}{{\text{e}}^{6}}
\end{aligned}$
(d)
$\begin{aligned}
{{\log }_{k}}81&=2 \\
{{k}^{2}}&=81 \\
k&=\pm \sqrt{81} \\
& =9\,\,\text{or}-9
\end{aligned}$
$k=-9$ (rej, as $k>0$)
$\therefore k=9$
(a)
(b)
(c)
(d)
(a)
$\begin{aligned}
\ln 4x&=5 \\
{{\log }_{\text{e}}}4x&=5 \\
4x&={{\text{e}}^{5}} \\
x&=\frac{1}{4}{{\text{e}}^{5}}
\end{aligned}$
(b)
$\begin{aligned}
{{\log }_{x}}16&=4 \\
{{x}^{4}}&=16 \\
x&=\pm \sqrt[4]{16} \\
x&=2\,\,\text{or}\,\,-2
\end{aligned}$
$x=-2$ (rej, as $x>0$)
$\therefore x=2$
(c)
$\begin{aligned}
\ln \left( 3x \right)&=6 \\
{{\log }_{\text{e}}}\left( 3x \right)&=6 \\
{{\text{e}}^{6}}&=3x \\
x&=\frac{1}{3}{{\text{e}}^{6}}
\end{aligned}$
(d)
$\begin{aligned}
{{\log }_{k}}81&=2 \\
{{k}^{2}}&=81 \\
k&=\pm \sqrt{81} \\
& =9\,\,\text{or}-9
\end{aligned}$
$k=-9$ (rej, as $k>0$)
$\therefore k=9$
Special Properties to Take Note
The logarithm of $1$ to the any base will always yield $0$.
${{\log }_{a}}1=0$
e.g.
| ${{\log }_{5}}1=0$ | ${{\log }_{17.5}}1=0$ | $\ln 1=0$ |
The logarithm of any number to the same base will always yield $1$.
${{\log }_{a}}a=1$
e.g.
| ${{\log }_{11}}11=1$ | ${{\log }_{20.5}}20.5=1$ | $\ln e=1$ |
Product Law and Its Proof
The product law for logarithms can be applied to simplify a logarithm of a product of many terms. The product law for logarithms states that the sum of the logs of two numbers is equal to the log of the product of two numbers.
${{\log }_{a}}b+{{\log }_{a}}c={{\log }_{a}}\left( b\times c \right)$
Important Notes:
- ${{\log }_{a}}(p+q)\ne \left( {{\log}_{a}}p \right)\left( {{\log}_{a}}q \right)$
- ${{\log }_{a}}(p+q)\ne {{\log}_{a}}p+{{\log}_{a}}q$
$\ln 2+\ln5=\ln10$
${{\log }_{3}}2+{{\log }_{3}}20={{\log }_{3}}40$
${{\log }_{2}}3+{{\log }_{2}}7+{{\log }_{2}}2={{\log }_{2}}42$
Practice Question 3
Simplify ${{\lg }_{6}}2+{{\lg }_{6}}3$ if possible.
Solution
Solution
$\begin{aligned}
& {{\lg }_{6}}2+{{\lg }_{6}}3 \\
& ={{\log }_{6}}\left( 2\times 3 \right) \\
& ={{\log }_{6}}6 \\
& =1 \\
\end{aligned}$
Quotient Law and Its Proof
The quotient law for logarithms is also often used when simplifying logs. The quotient law for logarithms states that the difference between the logs of two numbers is equal to the log of the quotient of two numbers.
${{\log }_{a}}b-{{\log }_{a}}c={{\log }_{a}}\left( \frac{b}{c} \right)$
Important Notes:
- ${{\log }_{a}}(p-q)\ne \frac{{{\log}_{a}}p}{{{\log}_{a}}q}$
- ${{\log }_{a}}(p+q)\ne {{\log}_{a}}p+{{\log}_{a}}q$
$\ln 2-\ln5=\ln\frac{2}{5}$
${{\log }_{3}}2-{{\log }_{3}}20={{\log }_{3}}\frac{1}{10}$
${{\log }_{2}}16-{{\log }_{2}}5+{{\log }_{2}}2={{\log }_{2}}\frac{16\times 2}{5}$
Practice Question 4
Simplify ${{\log }_{5}}25-{{\log }_{5}}5$ if possible.
Solution
Solution
$\begin{aligned}
& {{\log }_{5}}25-{{\log }_{5}}5 \\
& ={{\log }_{5}}\left( \frac{25}{5} \right) \\
& ={{\log }_{5}}5 \\
& =1 \\
\end{aligned}$
Power Law and Its Proof
The power law for logarithms can be applied for expanding or simplifying logarithms. The power law of logarithms allows us to move the exponent to the front of the logarithm if the logarithmic term has an exponent.
${{\log }_{a}}{{b}^{r}}=r{{\log }_{a}}b$
Important Notes:
${{\left( {{\log }_{a}}x \right)}^{3}}\ne 3{{\log }_{a}}x$
But ${{\left( {{\log }_{a}}x \right)}^{3}}=\left( {{\log }_{a}}x \right)\left( {{\log }_{a}}x \right)\left( {{\log }_{a}}x \right)$
${{\log }_{2}}{{5}^{x}}=x{{\log }_{2}}5$
${{\log }_{3}}\sqrt{2}=\frac{1}{2}{{\log }_{3}}2$
Practice Question 5
Evaluate each of the following without using the calculator.
(a)
${{\log }_{8}}64$
(a) ${{\log }_{8}}64$
(b)
${{\log }_{2}}\sqrt{4}+{{\log }_{2}}\sqrt{3}-{{\log }_{2}}\sqrt{6}$
(b) ${{\log }_{2}}\sqrt{4}+{{\log }_{2}}\sqrt{3}-{{\log }_{2}}\sqrt{6}$
(a)
(b)
(a)
$\begin{aligned}
{{\log }_{8}}64&={{\log }_{8}}{{8}^{2}} \\
& =2{{\log }_{8}}8 \\
& =2\left( 1 \right) \\
& =2
\end{aligned}$
(b)
$\begin{aligned}
& {{\log }_{2}}\sqrt{4}+{{\log }_{2}}\sqrt{3}-{{\log }_{2}}\sqrt{6} \\
& ={{\log }_{2}}\left( \sqrt{4}\cdot \sqrt{3} \right)-{{\log }_{2}}\sqrt{6} \\
& ={{\log }_{2}}\frac{\sqrt{4}\cdot \sqrt{3}}{\sqrt{6}} \\
& ={{\log }_{2}}\sqrt{\frac{4\cdot 3}{6}} \\
& ={{\log }_{2}}\sqrt{2} \\
& ={{\log }_{2}}{{2}^{\frac{1}{2}}} \\
& =\frac{1}{2}{{\log }_{2}}2 \\
& =\frac{1}{2} \\
\end{aligned}$
(a)
(b)
(a)
$\begin{aligned}
{{\log }_{8}}64&={{\log }_{8}}{{8}^{2}} \\
& =2{{\log }_{8}}8 \\
& =2\left( 1 \right) \\
& =2
\end{aligned}$
(b)
$\begin{aligned}
& {{\log }_{2}}\sqrt{4}+{{\log }_{2}}\sqrt{3}-{{\log }_{2}}\sqrt{6} \\
& ={{\log }_{2}}\left( \sqrt{4}\cdot \sqrt{3} \right)-{{\log }_{2}}\sqrt{6} \\
& ={{\log }_{2}}\frac{\sqrt{4}\cdot \sqrt{3}}{\sqrt{6}} \\
& ={{\log }_{2}}\sqrt{\frac{4\cdot 3}{6}} \\
& ={{\log }_{2}}\sqrt{2} \\
& ={{\log }_{2}}{{2}^{\frac{1}{2}}} \\
& =\frac{1}{2}{{\log }_{2}}2 \\
& =\frac{1}{2} \\
\end{aligned}$
Change of Base and Its Proof
From the previous section, we know that the common logarithm is also called the “logarithm with a base of $10$,” and the natural logarithm is referred to as the “logarithm with a base of $\text{e}$”. Although both logarithms are commonly called the logarithm, they are actually two separate calculations. Most graphing calculators have keys that allow you to work with logarithms: LOG for common logarithms and LN for natural logarithms.
What happens if we have questions involving logarithms with a base other than $10$ and $\text{e}$? The change-of-base formula allows us to convert any logarithm into another logarithm with a different base, which can be very useful in solving certain types of questions.
${{\log }_{a}}b=\frac{{{\log }_{c}}b}{{{\log }_{c}}a}$
Important Notes:
${{\log }_{x}}y=\frac{1}{{{\log }_{y}}x}$
Converting the logarithm into another logarithm with base $7$:
${{\log }_{2}}5=\frac{{{\log }_{7}}5}{{{\log }_{7}}2}$
Converting the logarithm into another logarithm with base $p$:
${{\log }_{11}}2=\frac{{{\log }_{p}}2}{{{\log }_{p}}11}$
Practice Question 6
(a)
Given that $p={{\log }_{a}}9$, find ${{\log }_{3}}a$ in terms of $p$.
(a) Given that $p={{\log }_{a}}9$, find ${{\log }_{3}}a$ in terms of $p$.
(b)
If $p=\lg 14$, ${{\log }_{14}}1\frac{2}{5}$ in terms of $p$.
(b) If $p=\lg 14$, ${{\log }_{14}}1\frac{2}{5}$ in terms of $p$.
(a)
(b)
(a)
$\begin{aligned}
p&={{\log }_{a}}9 \\
p&=\frac{{{\log }_{3}}9}{{{\log }_{3}}a} \\
{{\log }_{3}}a&=\frac{{{\log }_{3}}{{3}^{2}}}{p} \\
& =\frac{2{{\log }_{3}}3}{p} \\
& =\frac{2}{p}
\end{aligned}$
(b)
$\begin{aligned}
p&=\lg 14\\&={{\log }_{10}}14\\&=\frac{{{\log }_{14}}14}{{{\log }_{14}}10}\\&=\frac{1}{{{\log }_{14}}10} \\
{{\log }_{14}}10&=\frac{1}{p}
\end{aligned}$
$\begin{aligned}
& {{\log }_{14}}\left( 1\frac{2}{5} \right)\\&={{\log }_{14}}\left( \frac{7}{5} \right) \\
& ={{\log }_{14}}\left( \frac{7\cdot 2}{5\cdot 2} \right) \\
& ={{\log }_{14}}\left( \frac{14}{10} \right) \\
& ={{\log }_{14}}14-{{\log }_{14}}10 \\
& =1-\frac{1}{p}
\end{aligned}$
(a)
(b)
(a)
$\begin{aligned}
p&={{\log }_{a}}9 \\
p&=\frac{{{\log }_{3}}9}{{{\log }_{3}}a} \\
{{\log }_{3}}a&=\frac{{{\log }_{3}}{{3}^{2}}}{p} \\
& =\frac{2{{\log }_{3}}3}{p} \\
& =\frac{2}{p}
\end{aligned}$
(b)
$\begin{aligned}
p&=\lg 14\\&={{\log }_{10}}14\\&=\frac{{{\log }_{14}}14}{{{\log }_{14}}10}\\&=\frac{1}{{{\log }_{14}}10} \\
{{\log }_{14}}10&=\frac{1}{p}
\end{aligned}$
$\begin{aligned}
& {{\log }_{14}}\left( 1\frac{2}{5} \right)\\&={{\log }_{14}}\left( \frac{7}{5} \right) \\
& ={{\log }_{14}}\left( \frac{7\cdot 2}{5\cdot 2} \right) \\
& ={{\log }_{14}}\left( \frac{14}{10} \right) \\
& ={{\log }_{14}}14-{{\log }_{14}}10 \\
& =1-\frac{1}{p}
\end{aligned}$
Simultaneous Logarithm Equations
Suppose we have a system of equations that contains multiple logarithmic equations and multiple unknown variables, how do we address them? In this section, we will learn how to solve systems of logarithmic equations by substitution.
Practice Question 7
Solve the simultaneous equations.
$\begin{aligned}
{{\log }_{4}}x-{{\log }_{2}}y&=2 \\
{{3}^{x}}&=81\left( {{9}^{\frac{3}{2}-3y}} \right)
\end{aligned}$
Solution
Solution
$\begin{aligned}
{{\log }_{4}}x-{{\log }_{2}}y&=2—(1) \\
{{3}^{x}}&=81\left( {{9}^{\frac{3}{2}-3y}} \right)—(2)
\end{aligned}$
From (2):
$\begin{aligned}
{{3}^{x}}&={{3}^{4}}\left( {{\left( {{3}^{2}} \right)}^{\frac{3}{2}-3y}} \right) \\
{{3}^{x}}&={{3}^{4}}\cdot {{3}^{3-6y}} \\
{{3}^{x}}&={{3}^{4+3-6y}} \\
x&=4+3-6y \\
x&=7-6y—(3)
\end{aligned}$
Substitute (3) into (1):
$\begin{aligned}
{{\log }_{4}}\left( 7-6y \right)-{{\log }_{2}}y&=2 \\
\frac{{{\log }_{2}}\left( 7-6y \right)}{{{\log }_{2}}{{2}^{2}}}-{{\log }_{2}}y&=2 \\
\frac{{{\log }_{2}}\left( 7-6y \right)}{2}-{{\log }_{2}}y&=2 \\
{{\log }_{2}}\left( 7-6y \right)-2{{\log }_{2}}y&=4 \\
{{\log }_{2}}\left( 7-6y \right)-{{\log }_{2}}{{y}^{2}}&=4 \\
{{\log }_{2}}\frac{7-6y}{{{y}^{2}}}&=4 \\
\frac{7-6y}{{{y}^{2}}}&={{2}^{4}} \\
7-6y&=16{{y}^{2}} \\
16{{y}^{2}}+6y-7&=0\\
\left( 2y-1 \right)\left( 8y+7 \right)&=0 \\
y&=\frac{1}{2}\,\,\text{or}\,\,-\frac{7}{8}
\end{aligned}$
$y=-\frac{7}{8}$ (rej, as ${{\log }_{2}}\left( -\frac{7}{8} \right)$ is undefined)
Substitute $y=\frac{1}{2}$ into (3):
$\begin{aligned}
x&=7-6\left( \frac{1}{2} \right) \\
& =4
\end{aligned}$
$\therefore y=\frac{1}{2},\,\,x=4$
Applications of Logarithms
The logarithm is a very common and useful way to measure the relative magnitudes of things. Here are a few examples of logarithms in real-life:
- The Richter scale – a measure of the energy released in an earthquake, with each level 10 times more powerful than the last.
- Decibel – a logarithmic measure of the intensity of a sound.
- pH value – a measure of acidity and alkalinity on a scale of 0 to 14, with 7 being neutral.
Practice Question 8
The mass, $m$ grams, of a radioactive substance, present at time t days after first being observed, is given by the formula $m=28{{e}^{-0.00072t}}$.
(a)
Find the value of $m$ when $t=20$.
(a) Find the value of $m$ when $t=20$.
(b)
Find the value of $t$ when the mass is half of its value at $t=0$.
(b) Find the value of $t$ when the mass is half of its value at $t=0$.
(c)
State the value which $m$ approaches as $t$ becomes very large.
(c) State the value which $m$ approaches as $t$ becomes very large.
(d)
Sketch the graph of $m$ against $t$.
(d) Sketch the graph of $m$ against $t$.
(a)
(b)
(c)
(d)
(a)
When $t=20$,
$\begin{aligned}
m&=28{{\text{e}}^{-0.00072\left( 20 \right)}} \\
& =27.5996 \\
& =27.6\,\text{grams }\left( 3\,\,\text{s}\text{.f}\text{.} \right)
\end{aligned}$
(b)
When $t=0$,
$\begin{aligned}
m&=28{{\text{e}}^{-0.00072\left( 0 \right)}} \\
& =28{{\text{e}}^{0}} \\
& =28\,\text{grams}
\end{aligned}$
New $m=\frac{28}{2}=14$
$\begin{aligned}
m&=28{{\text{e}}^{-0.00072t}} \\
14&=28{{\text{e}}^{-0.00072t}} \\
\frac{1}{2}&={{\text{e}}^{-0.00072t}} \\
\ln \frac{1}{2}&=\ln {{\text{e}}^{-0.00072t}} \\
\ln \frac{1}{2}&=-0.00072t \\
t&=\frac{\ln \frac{1}{2}}{-0.00072} \\
& =962.7 \\
& =963\,\,\text{days}\,\left( 3\,\,\text{s}\text{.f}\text{.} \right)
\end{aligned}$
(c)
$t\to \infty ,m\to 0$
$m$ approaches $0$ gram.
(d)
Click image to view in full size
(a)
(b)
(c)
(d)
(a)
When $t=20$,
$\begin{aligned}
m&=28{{\text{e}}^{-0.00072\left( 20 \right)}} \\
& =27.5996 \\
& =27.6\,\text{grams }\left( 3\,\,\text{s}\text{.f}\text{.} \right)
\end{aligned}$
(b)
When $t=0$,
$\begin{aligned}
m&=28{{\text{e}}^{-0.00072\left( 0 \right)}} \\
& =28{{\text{e}}^{0}} \\
& =28\,\text{grams}
\end{aligned}$
New $m=\frac{28}{2}=14$
$\begin{aligned}
m&=28{{\text{e}}^{-0.00072t}} \\
14&=28{{\text{e}}^{-0.00072t}} \\
\frac{1}{2}&={{\text{e}}^{-0.00072t}} \\
\ln \frac{1}{2}&=\ln {{\text{e}}^{-0.00072t}} \\
\ln \frac{1}{2}&=-0.00072t \\
t&=\frac{\ln \frac{1}{2}}{-0.00072} \\
& =962.7 \\
& =963\,\,\text{days}\,\left( 3\,\,\text{s}\text{.f}\text{.} \right)
\end{aligned}$
(c)
$t\to \infty ,m\to 0$
$m$ approaches $0$ gram.
(d)
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