Logarithms

Matsec Syllabus: Pure Mathematics Paper 1 Topic 1: Logarithms:

Definition of logarithms;

The laws of logarithms;

Common and natural logarithms;

Change of base formula;

Solution of equations involving indices and logarithms.

Introduction

The following is the definition of the logarithm function:

The logarithm (or log, for short) of a number is a quantity representing the power (index/exponent) to which the base must be raised to produce that given number.

This definition means that an equation involving indices can be converted into an equation involving logs as follows:

\boxed{a^x=b\iff \log_a(b)=x} where b is a positive number.

Notes:

  1. The symbol \iff means that the two equations (in this case, a^x=b and \log_a(b)=x) are equivalent (so they describe the same thing).
  2. The number a of the logarithmic function \log_a (b) (written as a subscript) is known as the base of the logarithmic function.

Example 1: Convert the following equations into ones involving logs:

(a) 10^2=100

Substitute a=10, x=2, b=100 in the definition.

10^2=100\iff \log_{10}(100)=2

(b) 10^5=10000

Substitute a=10, x=5, b=10000 in the definition.

10^5=10000\iff \log_{10}(10000)=5

(c) 4^3=64

Substitute a=4, x=3, b=64 in the definition.

4^3=64\iff \log_4(64)=3

(d) e^{4.605}=100

Substitute a=e, x=4.605, b=100 in the definition.

e^{4.605}=100\iff \log_e(100)=4.605

Note that e is an irrational number whose value is approximately 2.71828182846

Example 2: Convert the following equations into ones involving indices:

(a) \log_3 81=4

Substitute a=3, x=4, b=81 in the definition.

\log_3 81=4\iff 3^4=81

 

(b) \log_{10} 10000=4

Substitute a=10, x=4, b=10000 in the definition.

\log_{10} 10000=4\iff 10^4=10000

(c) \log_4 2=\frac{1}{2}

Substitute a=4, x=\frac{1}{2}, b=2 in the definition.

\log_4 2=\frac{1}{2}\iff 4^{\frac{1}{2}}=2

(b) \log_e 1=0

Substitute a=e, x=0, b=1 in the definition.

\log_e 1=0\iff e^0=1

The Common & Natural Log

The logarithmic function with base 10, namely \log_10, is known as the common logarithm and is written as \log in short.

\boxed{\log_{10}=\log}

The logarithmic function with base e, namely \log_e, is known as the natural logarithm and is written as \ln in short. The letter “n” in “ln”, stands for natural.

\boxed{\log_e=\ln}

Example 3: Convert the following equations into ones involving indices:

(a) \log 100=2

\begin{aligned}\log 100&=2\\log_{10}100&=2\end{aligned}

Substitute a=10, x=2, b=100 in the definition.

\log_{10} 100=2\iff 10^2=100

 

(b) \ln e=1

\begin{aligned}\ln e&=1\\log_{e}e&=1\end{aligned}

Substitute a=e, x=1, b=e in the definition.

\log_{e} e=1\iff e^1=e

Example 4: Evaluate the following without using a calculator:

(a) \log_4 64

Let \log_4 64= x

We want to find the value of x.

\begin{aligned}\log_4 64 &=x\\4^x&=64\text{ (using the definition of logs)}\\4^x&=4^3\\\therefore x&=3\end{aligned}

(b) \log_{16} 2

Let \log_{16} 2= x

We want to find the value of x.

\begin{aligned}\log_{16} 2 &=x\\16^x&=2\text{ (using the definition of logs)}\\(2^4)^x&=2\\2^{4x}&=2\text{ (using Rule 3 of Indices)}\\\therefore 4x&=1\\x&=\frac{1}{4}\end{aligned}

(c) \log_a a^3

Let \log_a a^3= x

We want to find the value of x.

\begin{aligned}\log_a a^3&=x\\a^x&=a^3\text{ (using the definition of logs)}\\\therefore x&=3\end{aligned}

(d) \log 100000

Let \log_{10} 100000= x

We want to find the value of x.

\begin{aligned}\log_{10} 100000 &=x\\10^x&=100000\text{ (using the definition of logs)}\\10^x&=10^5\\\therefore x&=5\end{aligned}

Two Results of Logs

We have the following 2 results of logs that help us evaluate logs, more quickly without referring back to the indices.

Result 1 states that whenever the input of the log function is equal to its base, the output is always equal to 1.

Result 2 states that whenever the input of the log function is equal to 1, the output is always equal to 0.

Result 1:

\boxed{\log_a a=1} for any positive number a

Result 2:

\boxed{\log_a 1=0} for any positive number a

We are going to use these 2 results in order to evaluate logs.

Example 5: Evaluate the following without using a calculator:

(a) \log_5 5

\begin{aligned}&\log_5 5\\=&1\text{ (by Result 1)}\end{aligned}

(b) 20\log_8 8

\begin{aligned}&20\log_8 8\\=&20(1)\text{ (by Result 1)}\\=&20\end{aligned}

(c) \ln e

\begin{aligned}&\ln e\\=&\log_e e\\=&1\text{ (by Result 1)}\end{aligned}

(d) \log 10

\begin{aligned}&\log 10\\=&\log_{10} 10\\=&1\text{ (by Result 1)}\end{aligned}

(e) \log 1

\begin{aligned}&\log 1\\=&\log_{10} 1\\=&0\text{ (by Result 2)}\end{aligned}

(f) \ln 1

\begin{aligned}&\ln 1\\=&\log_{e} 1\\=&0\text{ (by Result 2)}\end{aligned}

(e) \frac{\log_4 1}{a}

\begin{aligned}&\frac{\log_4 1}{a}\\=&\frac{0}{a}\text{ (by Result 2)}\\=&0\end{aligned}

(f) \frac{1}{2}\ln 1

\begin{aligned}&\frac{1}{2}\ln 1\\=&\frac{1}{2}(0)\text{ (by Result 2)}\\=&0\end{aligned}

Three Rules of Logs

The following 3 rules of logs are very useful in simplifying and evaluating expressions involving logs.

Rule 1 states that the log of a product of two terms can be expressed as the sum of the log of the two terms.

Similarly, Rule 2 states that the log of a fraction of two terms can be expressed as the log of the numerator minus the log of the denominator.

Rule 3, states that when the input of the log function is in index form, the power can be removed from the input of the log function and added as a coeeficient of the log function.

Rule 1:

\boxed{\log_c{ab}=\log_c{a}+\log_c{b}} where a, b & c are positive numbers

Rule 2:

\boxed{\log_c{\frac{a}{b}}=\log_c{a}-\log_c{b}} where a, b & c are positive numbers

Rule 3:

\boxed{\log_c{a^b}=b\log_c{a}} where a, b & c are positive numbers

Let us now use these three rules in order to evaluate expressions involving logs.

Example 6: Simplify the following expressions:

(a) \log_3 a + \log_3 5

\begin{aligned}&\log_3 a + \log_3 5\\=&\log_3 5a\text{ (by Rule 1)}\end{aligned}

(b) \log 4x - \log x

\begin{aligned}&\log 4x – \log x\\=&\log(\frac{4x}{x})\text{ (by Rule 2)}\\=&\log 4\end{aligned}

(c) \log x - 1

\begin{aligned}&\log_x -1\\=&\log_x -\log 10\text{ (by Result 1)}\\=&\log(\frac{x}{10})\text{ (by Rule 2)}\end{aligned}

(d) \log 10 - \log 3+\log 60

\begin{aligned}&\log 10 – \log 3+\log 60\\=&\log(\frac{10}{3})\text{ (by Rule 2)}\\=&\log(\frac{10\times 60}{3})\text{ (by Rule 1)}\\=&\log 200\end{aligned}

(e) \log_2 10 - \log_2 6

\begin{aligned}&\log_2 10 – \log_2 6\\=&\log_2(\frac{10}{6})\text{ (by Rule 2)}\\=&\log_2(\frac{5}{3})\end{aligned}

(f) 2\log 4 - \log 2+3

\begin{aligned}&2\log 4 – \log 2+3\\=&2\log 4 – \log 2+3\log 10 \text{ (by Result 2)}\\=&\log 4^2 -\log 2 +\log 10^3 \text{ (by Rule 3)}\\=&\log 16 -\log 2 +\log 1000\\=&\log (\frac{16\times 1000}{2})\text{ (by Rules 1 and 2)}\\=&\log 8000\end{aligned}

(g) 4\log_3 2 - 2

\begin{aligned}&4\log_3 2 – 2\\=&4\log_3 2 – 2\log_3 3\text{ (by Result 1)}\\=&\log_3 2^4 – \log_3 3^2\text{ (by Rule 3)}\\=&\log_3 16 – \log_3 9\\=&\log_3(\frac{16}{9})\end{aligned}

(h) 2\ln a - \ln 2a

\begin{aligned}&2\ln a – \ln 2a\\=&\ln a^2 – \ln 2a \text{ (by Rule 3)}\\=&\ln (\frac{a^2}{2a})\text{ (by Rule 2)}\\=&\ln(\frac{a}{2})\end{aligned}

Example 7: Evaluate the following expressions:

(a) \log_4 64

\begin{aligned}&\log_4 64\\=&\log_4 4^3\\=&3\log_4 4\text{ (by Rule 3)}\\=&3(1)\text{ (by Result 1)}\\=&3\end{aligned}

(b) \frac{\log_3 16}{\log_3 64}

\begin{aligned}&\frac{\log_3 16}{\log_3 64}\\=&\frac{\log_3 2^4}{\log_3 2^6}\\=&\frac{4\log_3 2}{6\log_3 2}\text{ (by Rule 3)}\\=&\frac{4}{6}\\=&\frac{2}{3}\end{aligned}

Avoid the following common mistakes

Avoid the following common mistakes related to the use of the 3 Rules of Logs:

\begin{aligned} \bullet&\ \log_c (a+b) \neq \log_c a +\log_c b\\ \bullet&\ \log_c (a-b) \neq \log_c a -\log_c b\\ \bullet&\ \log_c (ab) \neq (\log_c a)(\log_c b)\\ \bullet&\ \log_c a^2 \neq (\log_c a)^2\\ \bullet&\ \log_c (\frac{a}{b}) \neq \frac{\log_c a}{\log_c b} \end{aligned} \begin{aligned} \qquad \end{aligned} \begin{aligned} \to&\text{ The correct rule is Rule 1: }\log_c (ab) = \log_c a +\log_c b\\ \to&\text{ The correct rule is Rule 2: }\log_c (\frac{a}{b}) = \log_c a -\log_c b\\ \to&\text{ The correct rule is Rule 1: }\log_c (ab) = \log_c a +\log_c b\\ \to&\text{ Using Rule 3, the correct equation is: }\log_c a^2 = 2\log_c a\\ \to&\text{ The correct rule is Rule 2: }\log_c (\frac{a}{b}) = \log_c a -\log_c b \end{aligned}

Change of Base Formula

The following is the formula that enables us to change the base of the log function. Here we are changing the base from b to c.

\log_b a = \frac{\log_c a}{\log_c b}

For example, suppose that we have \log_3 5 and we would like continue the working in base 4. Then by the Change of Base Formula:

\log_3 5 = \frac{\log_4 5}{\log_4 3}

Note that the Change of Base Formula was very handy in times when the calculator only evaluated log base 10 functions. Thus the Change of Base Formula was used mostly to convert a log of any base to base 10. Nowadays, our calculators and computers evaluate log functions of any base straight away, and thus this formula is mostly used to simply expressions and solving equations.

Example 7: Express the following with logs base 10 and evaluate:

(a) \log_7 49

\begin{aligned}&\log_7 49\\=&\frac{\log 49}{\log 7}\text{ (by Change of Base Formula)}\\=&\frac{\log 7^2}{\log 7}\\=&\frac{2\log 7}{\log 7}\text{ (by Rule 3)}\\=&2\end{aligned}

(b) \log_{11} 121

\begin{aligned}&\log_{11} 121\\=&\frac{\log 121}{\log 11}\text{ (by Change of Base Formula)}\\=&\frac{\log 11^2}{\log 11}\\=&\frac{2\log 11}{\log 11}\text{ (by Rule 3)}\\=&2\end{aligned}

(c) \log_{243} 9

\begin{aligned}&\log_{243} 9\\=&\frac{\log 9}{\log 243}\text{ (by Change of Base Formula)}\\=&\frac{\log 3^2}{\log 3^5}\\=&\frac{2\log 3}{5\log 3}\text{ (by Rule 3)}\\=&\frac{2}{5}\end{aligned}

(d) \log_4 2\sqrt{2}

\begin{aligned}&\log_{4} 2\sqrt{2}\\=&\frac{\log 2\sqrt{2}}{\log 4}\text{ (by Change of Base Formula)}\\=&\frac{\log 2^1 2^{\frac{1}{2}}}{\log 2^2}\\=&\frac{\log 2^{\frac{3}{2}}}{\log 2^2}\text{ (using the rules of indices)}\\=&\frac{\frac{3}{2}\log 2}{2\log 2}\text{ (by Rule 3)}\\=&\frac{\frac{3}{2}}{2}\\=&\frac{3}{4}\end{aligned}

(e) (\log_2 5)(\log_5 8)

\begin{aligned}&(\log_2 5)(\log_5 8)\\=&\frac{\log 5}{\log 2}\frac{\log 8}{\log 5}\text{ (by Change of Base Formula)}\\=&\frac{\log 8}{\log 2}\\=&\frac{\log 2^3}{\log 2}\\=&\frac{3\log 2}{\log 2}\text{ (by Rule 3)}\\=&3\end{aligned}

(f) \log_a b^2\log_b a^3

\begin{aligned}&\log_a b^2\log_b a^3\\=&\frac{\log b^2}{\log a}\frac{\log a^3}{\log b}\text{ (by Change of Base Formula)}\\=&\frac{2\log b}{\log a}\frac{3\log a}{\log b}\text{ (by Rule 3)}\\=&(2)(3)\\=&6\end{aligned}

Example 8: Use the Change of Base formula to show that: \log_b a =\frac{1}{\log_a b}

\begin{aligned}\log_b a &=\frac{\log_a a}{\log_a b}\text{ (by Change of Base formula)}\\&=\frac{1}{\log_a b}\text{ (by Result 1)}\end{aligned}

Solving indical equations by using logs

In the section on Indices we solved indical equations, that is, equations in which the unknown variable is in the power. For example, 2^x=16. We used the fact that 16=2^4. Hence we obtained both the left hand side and the right-hand side as an index having the same base. See worked example below.

Sometimes we can have an indical equation in which this is not possible, for example when we have 2^x=15 and thus we solve it by using logs. The trick is to put a log on each side of the equation and Rule 3, in order to bring down x from the power. See the worked examples, in Example 9.

\begin{aligned}2^x&=16\\2^x&=2^4\\x&=4\end{aligned}

\begin{aligned}2^x&=15\end{aligned}

Here we cannot express 15 in index form with base 2. Hence we must use logs.

Example 9: Solve the following equations:

(a) 2^x=15

\begin{aligned}2^x&=15\\\log 2^x&=\log 15\text{ (put log on both sides)}\\x\log 2&=\log 15\text{ (by Rule 3)}\\x&=\frac{\log 15}{\log 2}\\x&=3.91\end{aligned}

(b) 3^{x+1}=4

\begin{aligned}3^{x+1}&=4\\\log 3^{x+1}&=\log 4\text{ (put log on both sides)}\\(x+1)\log 3&=\log 4\text{ (by Rule 3)}\\x+1&=\frac{\log 4}{\log 3}\\x&=\frac{\log 4}{\log 3}-1\\x&=0.26\end{aligned}

(c) 2^x=3^{1-x}

\begin{aligned}2^x&=3^{1-x}\\\log 2^x&=\log 3^{1-x}\text{ (put log on both sides)}\\x\log 2&=(1-x)\log 3\text{ (by Rule 3)}\\x\log 2&=\log 3 -x\log 3\text{ (expanding R.H.S.*)}\\x\log 2 + x\log 3&=\log 3\\x(\log 2 +\log 3) &= \log 3\text{ (factorising L.H.S.)}\\x&=\frac{\log 3}{\log 2+ \log 3}\\x&=0.61\end{aligned}

* R.H.S. stands for right-hand-side (of the equation)

(d) (\frac{1}{9.8})^x=1.5\times 10^{-94.7}

\begin{aligned}(\frac{1}{9.8})^x&=1.5\times 10^{-94.7}\\\log(\frac{1}{9.8})^x&=\log (1.5\times 10^{-94.7})\text{ (put log on both sides)}\\x\log(\frac{1}{9.8})&=\log (1.5\times 10^{-94.7})\text{ (by Rule 3)}\\x\log(\frac{1}{9.8})&=\log 1.5+\log 10^{-94.7}\text{ (by Rule 1)}\\x\log(\frac{1}{9.8})&=\log 1.5-94.7\log 10\text{ (by Rule 3)}\\x\log(\frac{1}{9.8})&=\log 1.5-94.7(1)\text{ (by Result 1)}\\x&=\frac{\log 1.5-94.7}{\log(\frac{1}{9.8})}\\x&=95.36\end{aligned}