We think you are located in South Africa. Is this correct?

# Enrichment: More On Logarithms

## 2.8 Enrichment: more on logarithms (EMCFR)

NOTE: THIS SECTION IS NOT PART OF THE CURRICULUM

### Laws of logarithms (EMCFS)

Logarithmic law:

$\log_{a}{xy} = \log_{a}{x} + \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)$
\begin{align*} \text{Let } {\log}_{a}\left(x\right) = m & \quad \implies \quad x = a^{m} \ldots (1)\qquad (x > 0 )\\ \text{and }{\log}_{a}\left(y\right) = n & \quad \implies \quad x = a^{n} \ldots (2) \qquad (y > 0) \end{align*}\begin{align*} \text{Then } (1) \times (2): \quad x \times y &= a^{m} \times a^{n}\\ \therefore xy &= a^{m + n} \end{align*}

Now we change from the exponential form back to logarithmic form:

\begin{align*} \log_{a}{xy} &= m + n \\ \text{But } m = {\log}_{a}\left(x\right) & \text{ and } n = {\log}_{a}\left(y\right) \\ \therefore \log_{a}{xy} &= {\log}_{a}\left(x\right) +{\log}_{a}\left(y\right) \end{align*}

In words: the logarithm of a product is equal to the sum of the logarithms of the factors.

## Worked example 22: Applying the logarithmic law $$\log_{a}{xy} = \log_{a}{x} + \log_{a}{y}$$

Simplify: $$\log{5} + \log{2} - \log{30}$$

### Use the logarithmic law to simplify the expression

We combine the first two terms since the product of $$\text{5}$$ and $$\text{2}$$ is equal to $$\text{10}$$, which is always useful when simplifying logarithms.

\begin{align*} \log{5} + \log{2} - \log{30} &= \left( \log{5} + \log{2} \right) - \log{30} \\ &= \log{\left( 5 \times 2\right)} - \log{30} \\ &= \log{10} - \log{30} \\ &= 1 - \log{30} \end{align*}

We expand the last term to simplify the expression further:

\begin{align*} &= 1 - \log{( 3 \times 10)} \\ &= 1 - \left( \log{3} + \log{10} \right) \\ &= 1 - \left( \log{3} + 1 \right) \\ &= 1 - \log{3} - 1 \\ &= - \log{3} \end{align*}

$$\log{5} + \log{2} - \log{30} = - \log{3}$$

# Don't get left behind

Join thousands of learners improving their maths marks online with Siyavula Practice.

## Applying logarithmic law: $$\log_{a}{xy} = {\log}_{a}\left(x\right) +{\log}_{a}\left(y\right)$$

Exercise 2.14

Simplify the following, if possible:

$$\log_{8}{(10 \times 10)}$$

\begin{align*} \log_{8}{(10 \times 10)} &= \log_{8}{10} + \log_{8}{10} \\ &= 2 \log_{8}{10} \end{align*}

$$\log_{2}{14}$$

\begin{align*} \log_{2}{14} &= \log_{2}{(2 \times 7)} \\ &= \log_{2}{2} +\log_{2}{7} \\ &= 1 +\log_{2}{7} \end{align*}

$$\log_{2}{(8 \times 5)}$$

\begin{align*} \log_{2}{(8 \times 5)} &= \log_{2}{(2 \times 2 \times 2 \times 5)} \\ &= \log_{2}{2} + \log_{2}{2} +\log_{2}{2} + \log_{2}{5} \\ &= 3\log_{2}{2} +\log_{2}{5} \\ &= 3(1) +\log_{2}{5} \end{align*}

$$\log_{16}{(x + y)}$$

$$\log_{16}{(x + y)}$$ cannot be written as separate logarithms.

$$\log_{2}{2xy}$$

\begin{align*} \log_{2}{2xy} &= \log_{2}{(2 \times x \times y)} \\ &= \log_{2}{2} + \log_{2}{x} +\log_{2}{y} \\ &= 1 + \log_{2}{x} +\log_{2}{y} \end{align*}

$$\log{(5 + 2)}$$

$$\log{(5 + 2)} = \log{7}$$

Note: $$\log{(5 + 2)} \ne \log{5} + \log{2}$$. Do not confuse this with applying the distributive law: $$a(b + c) = ab + ac$$.

Write the following as a single term, if possible:

$$\log{15} + \log{2}$$

\begin{align*} \log{15} + \log{2} &= \log{(15 \times 2 )} \\ &= \log{30} \end{align*}

$$\log{1} + \log{5} +\log{\frac{1}{5}}$$

\begin{align*} \log{1} + \log{5} +\log{\frac{1}{5}} &= \log{\left( 1 \times 5 \times \frac{1}{5} \right) } \\ &= \log{1} \\ &= 0 \end{align*}

$$1 + \log_{3}{4}$$

\begin{align*} 1 + \log_{3}{4} &= \log_{3}{3} + \log_{3}{4} \\ &= \log_{3}{(3 \times 4) } \\ &= \log_{3}{12} \end{align*}

$$\left( \log{x} \right) \left( \log{y} \right) + \log{x}$$

\begin{align*} \left( \log{x} \right) \left( \log{y} \right) + \log{x} &= \left( \log{x} \right) [ \log{y} + 1 ] \\ &= \left( \log{x} \right) \left( \log{y} + \log{10} \right) \\ &= \left( \log{x} \right) \left( \log{10y} \right) \end{align*}

$$\log{7} \times \log{2}$$

\begin{align*} \log{7} \times \log{2} &= \log{7} \times \log{2} \\ & \text{cannot be simplified further} \\ \text{Note: } \quad \log{7} \times \log{2} &\ne \log{(7 + 2)} \end{align*}

$$\log_{2}{7} + \log_{3}{2}$$

This cannot be written as one term because the bases are not the same.

$$\log_{a}{p} + \log_{a}{q}$$

\begin{align*} \log_{a}{p} + \log_{a}{q} &= \log_{a}{(p \times q)} \\ &= \log_{a}{pq} \end{align*}

$$\log_{a}{p} \times \log_{a}{q}$$

This is already a single term: $$(\log_{a}{p})(\log_{a}{q})$$

Simplify the following:

$$\log{x} + \log{y} + \log{z}$$

$\log_{x} + \log{y} + \log{z} = \log{xyz}$

$$\log{ab} + \log{bc} + \log{cd}$$

$\log{ab} + \log{bc} + \log{cd} = \log{ab^{2}c^{2}d}$

$$\log{125} + \log{2} + \log{8}$$

\begin{align*} \log{125} + \log{2} + \log{8} &= \log{(125 \times 2 \times 8) } \\ &= \log{2000} \\ &= \log{(2 \times 10 \times 10 \times 10)} \\ &= \log{2} + \log{10} + \log{10} + \log{10} \\ &= \log{2} + 1 + 1 +1 \\ &= \log{2} + 3 \end{align*}

$$\log_{4}{\frac{3}{8}} + \log_{4}{\frac{10}{3}} + \log_{4}{\frac{16}{5}}$$

\begin{align*} \log_{4}{\frac{3}{8}} + \log_{4}{\frac{10}{3}} + \log_{4}{\frac{16}{5}} &= \log_{4}{\left( \frac{3}{8} \times \frac{10}{3} \times \frac{16}{5} \right) } \\ &= \log_{4}{4}\\ &= 1 \end{align*}

Logarithmic law:

$\log_{a}{\frac{x}{y}} = \log_{a}{x} - \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)$
\begin{align*} \text{Let } {\log}_{a}\left(x\right) = m & \quad \implies \quad x = a^{m} \ldots (1) \qquad (x > 0)\\ \text{and }{\log}_{a}\left(y\right) = n & \quad \implies \quad y = a^{n} \ldots (2) \qquad (y > 0) \end{align*}\begin{align*} \text{Then } (1) \div (2): \quad \frac{x}{y} &= \frac{a^{m}}{a^{n}} \\ \therefore \frac{x}{y} &= a^{m - n} \end{align*}

Now we change from the exponential form back to logarithmic form:

\begin{align*} \log_{a}{\frac{x}{y}} &= m - n \\ \text{But } m = {\log}_{a}\left(x\right) & \text{ and } n = {\log}_{a}\left(y\right) \\ \therefore \log_{a}{\frac{x}{y}} &= {\log}_{a}\left(x\right) - {\log}_{a}\left(y\right) \end{align*}

In words: the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator.

## Worked example 23: Applying the logarithmic law $$\log_{a}{\frac{x}{y}} = \log_{a}{x} - \log_{a}{y}$$

Simplify: $$\log{40} - \log{4} + \log_{5}{\frac{8}{5}}$$

### Use the logarithmic law to simplify the expression

We combine the first two terms since both terms have the same base and the quotient of $$\text{40}$$ and $$\text{4}$$ is equal to $$\text{10}$$:

\begin{align*} \log{40} - \log{4} + \log_{5}{\frac{8}{5}} &= \left(\log{40} - \log{4} \right) + \log_{5}{\frac{8}{5}} \\ &= \left(\log{\frac{40}{4}} \right) + \log_{5}{\frac{8}{5}} \\ &= \log{10} + \log_{5}{\frac{8}{5}} \\ &= 1 + \log_{5}{\frac{8}{5}} \end{align*}

We expand the last term to simplify the expression further:

\begin{align*} &= 1 + \left( \log_{5}{8} - \log_{5}{5} \right) \\ &= 1 + \log_{5}{8} - 1 \\ &= \log_{5}{8} \end{align*}

$$\log{40} - \log{4} + \log_{5}{\frac{8}{5}} = \log_{5}{8}$$

# Don't get left behind

Join thousands of learners improving their maths marks online with Siyavula Practice.

## Applying logarithmic law: $$\log_{a}{\frac{x}{y}} = {\log}_{a}x - {\log}_{a}y$$

Exercise 2.15

Expand and simplify the following:

$$\log{\frac{100}{3}}$$
\begin{align*} \log{\frac{100}{3}} &= \log{100} - \log{3} \\ &= \log{(10 \times 10)} - \log{3} \\ &= \log{10} + \log{10} - \log{3} \\ &= 1 + 1 - \log{3} \\ &= 2 - \log{3} \end{align*}
$$\log_{2}{7\frac{1}{2}}$$
\begin{align*} \log_{2}{7\frac{1}{2}} &= \log_{2}{\frac{15}{2}} \\ &= \log_{2}{15} - \log_{2}{2} \\ &= \log_{2}{15} - 1 \end{align*}
$$\log_{16}{\frac{x}{y}}$$
$\log_{16}{\frac{x}{y}} = \log_{16}{x} - \log_{16}{y}$
$$\log_{16}{(x - y)}$$

This cannot be simplified.

$$\log_{5}{\frac{5}{8}}$$
\begin{align*} \log_{5}{\frac{5}{8}} &= \log_{5}{5} - \log_{5}{8} \\ &= 1 - \log_{5}{8} \end{align*}
$$\log_{x}{\frac{y}{r}}$$
$\log_{x}{\frac{y}{r}} = \log_{x}{y} - \log_{x}{r}$

Write the following as a single term:

$$\log{10} - \log{50}$$
\begin{align*} \log{10} - \log{50} &= \log{\frac{10}{50}} \\ &= \log{\frac{1}{5}} \\ &= \log{5^{-1}} \\ &= - \log{5} \end{align*}
$$\log_{3}{36} - \log_{3}{4}$$
\begin{align*} \log_{3}{36} - \log_{3}{4} &= \log_{3}{\frac{36}{4}} \\ &= \log_{3}{9} \\ &= \log_{3}{(3 \times 3) } \\ &= \log_{3}{3} + \log_{3}{3} \\ &= 1 + 1 \\ &= 2 \end{align*}
$$\log_{a}{p} - \log_{a}{q}$$
$\log_{a}{p} - \log_{a}{q} = \log_{a}{\frac{p}{q}}$
$$\log_{a}{(p - q)}$$

This cannot be simplified.

$$\log{15} - \log_{2}{5}$$

This cannot be simplified because the bases are not the same.

$$\log{15} - \log{5}$$
\begin{align*} \log{15} - \log{5} &= \log{\frac{15}{5}} \\ &= \log{3} \end{align*}

Simplify the following:

$$\log{450} - \log{9} - \log{5}$$
\begin{align*} \log{450} - \log{9} - \log{5} &= \log{ \left(\frac{450}{9 \times 5} \right) } \\ &= \log{10} \\ &= 1 \\ \text{Alternative method: } & \\ \log{450} - \log{9} - \log{5} &= \log{\frac{450}{9}} - \log{5} \\ &= \log{50} - \log{5} \\ &= \log{\frac{50}{5}}\\ &= \log{10} \\ &= 1 \end{align*}
$$\log{\frac{4}{5}} - \log{\frac{3}{25}} - \log{\frac{1}{15}}$$
\begin{align*} \log{\frac{4}{5}} - \log{\frac{3}{25}} - \log{\frac{1}{15}} &= \log{ \left( \frac{\frac{4}{5}}{ \frac{3}{25} \times \frac{1}{15} } \right) }\\ &= \log{ \left( \frac{\frac{4}{5}}{ \frac{1}{125} } \right) } \\ &= \log{100} \\ &= 2 \\ \text{Alternative method: } & \\ \log{\frac{4}{5}} - \log{\frac{3}{25}} - \log{\frac{1}{15}} &= \log{\frac{4}{5}} - \left( \log{\frac{3}{25}} + \log{\frac{1}{15}} \right) \\ &= \log{\frac{4}{5}} - \log{ \left( \frac{3}{25} \times \frac{1}{15} \right) } \\ &= \log{\frac{4}{5}} - \log{ \left( \frac{3}{25 \times 15} \right) } \\ &= \log{\frac{4}{5}} - \log{ \left( \frac{1}{125} \right) } \\ &= \log{ \left( \frac{4}{5} \div \frac{1}{125} \right) } \\ &= \log{ \left( \frac{4}{5} \times \frac{125}{1} \right) } \\ &= \log{ 100} \\ &= \log{10} + \log{10} \\ &= 1 + 1 \\ &= 2 \end{align*}

Vini and Dirk complete their mathematics homework and check each other's answers. Compare the two methods shown below and decide if they are correct or incorrect:

Question:

Simplify the following:

$\log{m} - \log{n} - \log{p} - \log{q}$

\begin{align*} \log{m} - \log{n} - \log{p} - \log{q} &= \left( \log{m} - \log{n} \right) - \log{p} - \log{q} \\ &= \left( \log{\frac{m}{n}} - \log{p} \right) - \log{q} \\ &= \log{\left( \frac{m}{n} \times \frac{1}{p} \right)} - \log{q} \\ &= \log{\frac{m}{np} } - \log{q} \\ &= \log{\frac{m}{np} \times \frac{1}{q}} \\ &= \log{\frac{m}{npq} } \end{align*}

\begin{align*} \log{m} - \log{n} - \log{p} - \log{q} &= \log{m} - \left( \log{n} + \log{p} + \log{q} \right) \\ &= \log{m} - \log{(n \times p \times q)} \\ &= \log{m} - \log{(npq)} \\ &= \log{\frac{m}{npq} } \end{align*}
Both methods are correct.

Useful summary:

1. $$\log1=0$$

2. $$\log10=1$$

3. $$\log100=2$$

4. $$\log1000=3$$

5. $$\log{\frac{1}{10}}=-1$$

6. $$\log{\text{0,1}}=-1$$

7. $$\log{\text{0,01}}=-2$$

8. $$\log{\text{0,001}}=-3$$

## Worked example 24: Simplification of logarithms

Simplify (without a calculator): $$3\log3+\log125$$

### Apply the appropriate logarithmic laws to simplify the expression

\begin{align*} 3\log3+\log125 &= 3\log3+\log{5^{3}} \\ &= 3\log3 + 3\log{5} \\ &= 3 \left( \log3 + \log{5} \right) \\ &= 3 \log{(3 \times 5)} \\ &= 3 \log{15} \end{align*}

We cannot simplify any further, therefore $$3\log3+\log125 = 3 \log{15}$$.

Important: all the algebraic manipulation techniques $$(\times, \div, +, -$$, factorisation etc.) also apply for logarithmic expressions. Always be aware of the number of terms in an expression as this will help to determine how to simplify.

## Simplification of logarithms

Exercise 2.16

Simplify the following without using a calculator:

$$8^{\frac{2}{3}} + \log_{2}{32}$$
\begin{align*} 8^{\frac{2}{3}} + \log_{2}{32} &= \left( 2^{3} \right)^{\frac{2}{3}} + \log_{2}{(2^{5})} \\ &= 2^{2} + 5 \log_{2}{2} \\ &= 4 + 5 \\ &= 9 \end{align*}
$$2 \log{3} + \log{2} - \log{5}$$
\begin{align*} 2 \log{3} + \log{2} - \log{5} &= \log{3^{2}} + \log{2} - \log{5} \\ &= \log{\frac{9 \times 2}{5}} \\ &= \log{\frac{18}{5}} \end{align*}
$$\log_{2}{8} - \log{1} + \log_{4}{\frac{1}{4}}$$
\begin{align*} \log_{2}{8} - \log{1} + \log_{4}{\frac{1}{4}} &= \log_{2}{2^{3}} - 0 + \log_{4}{4^{(-1)}} \\ &= 3 \log_{2}{2} - 1 \log_{4}{4} \\ &= 3(1) - 1(1) \\ &= 2 \end{align*}
$$\log_{8}{1} - \log_{5}{\frac{1}{25}} + \log_{3}{9}$$
\begin{align*} \log_{8}{1} - \log_{5}{\frac{1}{25}} + \log_{3}{9} &= 0 - \log_{5}{5^{(-2)}} + \log_{3}{3^{2}}\\ &= -(-2) \log_{5}{5} + 2 \log_{3}{3} \\ &= 2(1) + 2(1) \\ &= 4 \end{align*}

## Worked example 25: Solving logarithmic equations

Solve for $$p$$:

$18 \log{p} - 36 = 0$

### Make $$\log p$$ the subject of the equation

\begin{align*} 18 \log{p} - 36 &= 0 \\ 18 \log{p} &= 36 \\ \frac{18 \log{p}}{18} &= \frac{36}{18} \\ \therefore \log{p} &= 2 \end{align*}

### Change from logarithmic form to exponential form

\begin{align*} \log{p} &= 2 \\ \therefore p &= 10^{2} \\ &= 100 \end{align*}

$$p = 100$$

## Worked example 26: Solving logarithmic equations

Solve for $$n$$ (correct to the nearest integer):

$(\text{1,02})^{n} = 2$

### Change from exponential form to logarithmic form

\begin{align*} (\text{1,02})^{n} &= 2 \\ \therefore n &= \log_{\text{1,02}}{2} \end{align*}

### Use a change of base to solve for $$n$$

\begin{align*} n &= \frac{\log{2}}{\log{\text{1,02}}} \\ \therefore n &= \text{35,00} \ldots \end{align*}

$$n = 35$$

## Solving logarithmic equations

Exercise 2.17

Determine the value of $$a$$ (correct to $$\text{2}$$ decimal places):

$$\log_{3}{a} - \log{\text{1,2}} = 0$$
\begin{align*} \log_{3}{a} - \log{\text{1,2}} &= 0 \\ \log_{3}{a} &= \log{\text{1,2}} \\ \text{Change to exponential form: } & \\ 3^{\log{\text{1,2}}} &= a \\ \therefore a &= \text{1,09} \end{align*}

Alternative (longer) method:

\begin{align*} \log_{3}{a} - \log{\text{1,2}} &= 0 \\ \log_{3}{a} &= \log{\text{1,2}} \\ \frac{\log{a}}{\log{3}} &= \log{\text{1,2}} \\ \log{a} &= \log{3} \times \log{\text{1,2}} \\ \log{a} &= \text{0,037} \ldots \\ \therefore a &= \text{1,09} \end{align*}
$$\log_{2}{(a - 1)} = \text{1,5}$$
\begin{align*} \log_{2}{(a - 1)} &= \text{1,5} \\ \text{Change to exponential form: } & \\ 2^{\text{1,5}} &= a - 1 \\ 2^{\text{1,5}} + 1 &= a \\ \therefore a &= \text{3,83} \end{align*}

Alternative (longer) method:

\begin{align*} \log_{2}{(a - 1)} &= \text{1,5} \\ \frac{\log{(a - 1)}}{\log{2}} &= \text{1,5} \\ \log{(a - 1)} &= \log{2} \times \text{1,5} \\ \therefore a - 1 &= \text{2,83} \ldots \\ \therefore a &= \text{3,83} \end{align*}
$$\log_{2}{a} - 1 = \text{1,5}$$
\begin{align*} \log_{2}{a - 1} &= \text{1,5} \\ \log_{2}{a} &= \text{2,5} \\ \text{Change to exponential form: } & \\ 2^{\text{2,5}} &= a \\ \therefore a &= \text{5,66} \end{align*}

Alternative (longer) method:

\begin{align*} \log_{2}{a} - 1 &= \text{1,5} \\ \frac{\log{a}}{\log{2}} &= \text{2,5} \\ \log{a} &= \log{2} \times \text{2,5} \\ \therefore a &= \text{5,66} \end{align*}
$$3^{a} = \text{2,2}$$
\begin{align*} 3^{a} &= \text{2,2} \\ \therefore a &= \log_{3}{\text{2,2}}\\ &= \frac{\log{\text{2,2}}}{\log{3}} \\ \therefore a &= \text{0,72} \end{align*}
$$2^{(a + 1)} = \text{0,7}$$
\begin{align*} 2^{(a + 1)} &= \text{0,7} \\ \therefore a + 1 &= \log_{2}{\text{0,7}}\\ \therefore a &= \frac{\log{\text{0,7}}}{\log{2}} - 1 \\ &= -\text{1,51} \end{align*}
$$(\text{1,03})^{\frac{a}{2}} = \text{2,65}$$
\begin{align*} (\text{1,03})^{\frac{a}{2}} &= \text{2,65}\\ \therefore \frac{a}{2}&= \log_{\text{1,03}}{\text{2,65}}\\ \therefore a &= 2 \times \frac{\log{\text{2,65}}}{\log{\text{1,03}}}\\ &= \text{65,94} \end{align*}
$$(\text{9})^{(1 - 2a)} = \text{101}$$
\begin{align*} (\text{9})^{(1 - 2a)} &= \text{101}\\ \therefore 1 - 2a &= \log_{\text{9}}{\text{101}} \\ \therefore 1 - \frac{\log{\text{101}}}{\log{\text{9}}} &= 2a \\ -\text{1,10} \ldots &= 2a \\ \therefore -\text{0,55} &= a \end{align*}

Given $$y = 3^{x}$$.

Write down the equation of the inverse of $$y = 3^x$$ in the form $$y = \ldots$$
$$y = \log_{3}{x}$$
If $$6 = 3^{p}$$, determine the value of $$p$$ (correct to one decimal place).
\begin{align*} p &= \log_{3}{6} \\ &= \frac{\log{6}}{\log{3}} \\ &=\text{1,6} \end{align*}
Draw the graph of $$y= 3^{x}$$ and its inverse. Plot the points $$A(p; 6)$$ and $$B(6; p)$$.

### Summary (EMCFW)

• The logarithm of a number $$(x)$$ with a certain base $$(a)$$ is equal to the exponent $$(y)$$, the value to which that certain base must be raised to equal the number $$(x)$$.

If $$x = {a}^{y}$$, then $$y = {\log}_{a}\left(x\right)$$, where $$a>0$$, $$a \ne 1$$ and $$x>0$$.

• Logarithms and exponentials are inverses of each other.

$$f(x) = \log_{a}{x} \quad$$ and $$\quad f^{-1}(x) = a^{x}$$

• Common logarithm: $$\log{a}$$ means $$\log_{\text{10}}{a}$$
• The “LOG” function on your calculator uses a base of $$\text{10}$$.
• Natural logarithm: $$\ln$$ uses a base of $$e$$.
• Special values:

• $$a^{0} = 1 \qquad \log_{a}{1} = 0$$
• $$a^{1} = a \qquad \log_{a}{a} = 1$$
• Logarithmic laws:

• $$\log_{a}{xy} = \log_{a}{x} + \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)$$
• $$\log_{a}{\frac{x}{y}} = \log_{a}{x} - \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)$$
• $$\log_{a}{x^{b}} = b \log_{a}{x} \qquad (x > 0)$$
• $$\log_{a}{x} = \frac{\log_{b}{x}}{\log_{b}{a}} \qquad (b > 0 \text{ and } b \ne 1)$$
• Special reciprocal applications:

• $$\log_{a}{x} = \frac{1}{\log_{x}{a}}$$
• $$\log_{a}{\frac{1}{x}} = - \log_{a}{x}$$

## Logarithms (ENRICHMENT ONLY)

Exercise 2.18

State whether the following are true or false. If false, change the statement so that it is true.

$$\log{t} + \log{d} = \log{(t + d)}$$

False: $$\log{t} + \log{d} = \log{(t \times d)}$$

If $$p^{q} = r$$, then $$q = \log_{r}{p}$$

False: $$q = \log_{p}{r}$$

$$\log{\frac{A}{B}} = \log{A} - \log{B}$$

True

$$\log{A} - B = \frac{\log{A}}{\log{B}}$$

False: $$\log{(A)} - B$$ cannot be simplified further.

$$\log_{\frac{1}{2}}{x} = - \log_{2}{x}$$

True

$$\log_{k}{m} = \frac{\log_{p}{k}}{\log_{p}{m}}$$

False: $$\log_{k}{m} = \frac{\log_{p}{m}}{\log_{p}{k}}$$

$$\log_{n}{\sqrt{b}} = \frac{1}{2}\log_{n}{b}$$

True

$$\log_{p}{q} = \frac{1}{\log_{q}{p}}$$

True

$$2\log_{2}{a} + 3 \log{a} = 5 \log{a}$$

False: bases are different

$$5\log{x} + 10 \log{x} = 5 \log{x^{3}}$$

True

$$\frac{\log_{n}{a}}{\log_{n}{b}} = \log_{n}{\frac{a}{b}}$$

False: cannot be simplified to single logarithm

$$\log{(A + B)} = \log{A} + \log{B}$$

False: do not confuse with $$\log{(AB)} = \log{A} + \log{B}$$ or with the distributive law $$x(a + b) = ax + ab$$.

$$\log{2a^{3}} = 3 \log{2a}$$

False: $$\log{2a^{3}} = \log{2} + 3\log{a}$$

$$\frac{\log_{n}{a}}{\log_{n}{b}} = \log_{n}{(a - b)}$$

False: do not confuse with $$\log_{n}{ \left( \frac{a}{b} \right) } = \log_{n}{a} - \log_{n}{b}$$. LHS cannot be simplified.

Simplify the following without using a calculator:

$$\log{7} - \log{\text{0,7}}$$
\begin{align*} \log{7} - \log{\text{0,7}} &= \log{\frac{\text{7}}{\text{0,7}}} \\ &= \log{\text{10}} \\ &= \text{1} \end{align*}
$$\log{8} \times \log{\text{1}}$$
\begin{align*} \log{8} \times \log{\text{1}} &= \log{8} \times \text{0}\\ &= \text{0} \end{align*}
$$\log{\frac{1}{3}} + \log{\text{300}}$$
\begin{align*} \log{\frac{1}{3}} + \log{\text{300}} &= \log{\left( \frac{1}{3} \times \text{300} \right) }\\ &= \log{\text{100}} \\ &= \log{\text{10}^{2}} \\ &= 2 \log{\text{10}} \\ &= \text{2} \end{align*}
$$2\log{3} + \log{2} - \log{6}$$
\begin{align*} 2\log{3} + \log{2} - \log{6} &= \log{3^{2}} + \log{2} - \log{6} \\ &= \log{\frac{9 \times 2}{6}} \\ &= \log{\frac{18}{6}} \\ &= \log{3} \end{align*}

Given $$\log{5} = \text{0,7}$$. Find the value of the following without using a calculator:

$$\log{50}$$
\begin{align*} \log{50} &= \log{5} + \log{10} \\ &= \text{0,7} + 1 \\ &= \text{1,7} \end{align*}
$$\log{20}$$
\begin{align*} \log{20} &= \log{\frac{100}{5}} \\ &= \log{100} - \log{5} \\ &= 2 - \text{0,7} \\ &= \text{1,3} \end{align*}
$$\log{25}$$
\begin{align*} \log{25} &= \log{5^{2}} \\ &= 2 \times \log{5} \\ &= 2 \times \text{0,7} \\ &= \text{1,4} \end{align*}
$$\log_{2}{5}$$
\begin{align*} \log_{2}{5} &= \frac{\log{5}}{\log{2}} \\ &= \frac{\log{5}}{\log{\frac{10}{5}}} \\ &= \frac{\log{5}}{\log{10} - \log{5}} \\ &= \frac{\text{0,7}}{1 - \text{0,7}} \\ &= \frac{\text{0,7}}{\text{0,3}} \\ &= \frac{7}{10} \times \frac{10}{3} \\ &= \frac{7}{3} \end{align*}
$$10^{\text{0,7}}$$

If $$\log{5} = \text{0,7}$$ , then $$10^{\text{0,7}} = \text{5}$$.

Given $$A = \log_{8}{1} - \log_{5}{\frac{1}{25}} + \log_{3}{9}$$.

Without using a calculator, show that $$A = 4$$.
\begin{align*} A &= \log_{8}{1} - \log_{5}{\frac{1}{25}} + \log_{3}{9} \\ &= 0 - \log_{5}{5^{-2}} + \log_{3}{3^{2}} \\ &= 0 - (-2)\log_{5}{5} + (2)\log_{3}{3} \\ &= 2 + 2 \\ &= 4 \end{align*}
Now solve for $$x$$ if $$\log_{2}{x} = A$$.
\begin{align*} \log_{2}{x} &= 4 \\ \therefore x &= 2^{4} \\ x &= 16 \end{align*}
Let $$f(x) = \log_{2}{x}$$. Draw the graph of $$f$$ and $$f^{-1}$$. Indicate the point $$(x;A)$$ on the graph.

Solve for $$x$$ if $$\frac{35^{x}}{7^{x}} = \text{15}$$. Give answer correct to two decimal places.

\begin{align*} \frac{35^{x}}{7^{x}} &= \text{15} \\ \frac{7^{x} \cdot 5^{x}}{7^{x}} &= \text{15} \\ 5^{x} &= \text{15} \\ \therefore x &= \log_{5}{15} \\ &= \frac{\log{15}}{\log{5}} \\ &= \text{1,68} \end{align*}

Given $$f(x) = 5 \times \left( \text{1,5} \right)^{x}$$ and $$g(x) = \left( \frac{1}{4} \right)^{x}$$.

For which integer values of $$x$$ will $$f(x) < \text{295}$$.
\begin{align*} 5 \times \left( \text{1,5} \right)^{x} &< \text{295} \\ \therefore \log{(\text{1,5})^{x}} &< \log{59} \\ x \log{(\text{1,5})} &< \log{59} \\ x &< \frac{ \log{59}}{\log{(\text{1,5})} } \qquad \text{note: } \log{(\text{1,5})} > 0 \\ x &< \text{10,0564} \ldots \end{align*}

Therefore, $$x < 10, (x \in \mathbb{Z})$$.

For which values of $$x$$ will $$g(x) \geq \text{2,7} \times \text{10}^{-\text{7}}$$. Give answer to the nearest integer.
\begin{align*} \left( \frac{1}{4} \right)^{x} & \geq \text{2,7} \times \text{10}^{-\text{7}} \\ \log{\left( \frac{1}{4} \right)^{x}} & \geq \log{\text{2,7} \times \text{10}^{-\text{7}}} \\ x \log{\left( \frac{1}{4} \right)} & \geq \log{\text{2,7} \times \text{10}^{-\text{7}}} \\ x & \leq \frac{\log{\text{2,7} \times \text{10}^{-\text{7}}}}{\log{\left( \frac{1}{4} \right)}} \qquad \text{note: } \log{\left( \frac{1}{4} \right)} < 0 \\ & \leq \text{10,9} \ldots \\ \therefore x & < \text{11} \end{align*}

Important: notice that the inequality sign changed direction when we divided both sides by $$\log{\left( \frac{1}{4} \right)} = -\log{4}$$, since it has a negative value.