2.8 Enrichment: more on logarithms

3.1 Calculating the period of an investment

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*}

Write the final answer

\(\log{5} + \log{2} - \log{30} = - \log{3}\)

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

Textbook Exercise 2.14

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

\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: \(a(b + c) = ab + ac\).

\(\log{15} + \log{2}\)

\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: \((\log_{a}{p})(\log_{a}{q})\)

\(\log{x} + \log{y} + \log{z}\)

\[\log_{x} + \log{y} + \log{z} = \log{xyz}\]

\(\log{ab} + \log{bc} + \log{cd}\)

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

\(\log{125} + \log{2} + \log{8}\)

\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*}

Write the final answer

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

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

Textbook Exercise 2.15

\(\log{\frac{100}{3}}\)

\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*}

\(\log_{x}{\frac{y}{r}}\)

\[\log_{x}{\frac{y}{r}} = \log_{x}{y} - \log_{x}{r}\]

\(\log{10} - \log{50}\)

\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*}

\(\log{450} - \log{9} - \log{5}\)

\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:

\[\log{m} - \log{n} - \log{p} - \log{q}\]

Vini's answer:

\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*}

Dirk's answer:

\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:

\(\log1=0\)
\(\log10=1\)
\(\log100=2\)
\(\log1000=3\)
\(\log{\frac{1}{10}}=-1\)
\(\log{\text{0,1}}=-1\)
\(\log{\text{0,01}}=-2\)
\(\log{\text{0,001}}=-3\)

Simplification of logarithms (EMCFT)

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*}

Write the final answer

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

Textbook Exercise 2.16

\(8^{\frac{2}{3}} + \log_{2}{32}\)

\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*}

Solving logarithmic equations (EMCFV)

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*}

Write the final answer

\(p = 100\)

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*}

Write the final answer

\(n = 35\)

Solving logarithmic equations

Textbook Exercise 2.17

\(\log_{3}{a} - \log{\text{1,2}} = 0\)

\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*}

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)

Textbook Exercise 2.18

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

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.

\(\log{7} - \log{\text{0,7}}\)

\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*}

\(\log{50}\)

\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 \(10^{\text{0,7}} = \text{5}\).

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*}

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.

2.7 Summary

Table of Contents

3.1 Calculating the period of an investment