### Exponential and logarithmic functions - (3.1-3.3)

#### Understand the definitions of rational indices

• For any real numbers a and rational numbers m and n, the following arithmetic operations apply:

1. $$\sqrt[n]{a} = {a^{\frac{1}{n}}}$$

2. $${{\rm{a}}^{\frac{{\rm{m}}}{{\rm{n}}}}} = \sqrt[n]{{{a^m}}} = {\left( {\sqrt[n]{a}} \right)^m}$$

• Simplify the following expressions, and express your answers with positive indices.

a) $${x^{ - 2}} \times {x^{ - 3}}$$
b) $${(\frac{{{x^{ - 2}}}}{x})^{ - 2}}$$
c) $$\frac{{{{({x^{ - 1}})}^0}}}{{({x^{ - 2}})({x^4})}}$$
d) $${({x^{ - 1}}y)^2}{(2x{y^{ - 1}})^{ - 2}}$$
• $$\begin{array}{1}a)\qquad{x^{ - 2}} \times {x^{ - 3}}\\=\frac{1}{{{x^2}}} \times \frac{1}{{{x^3}}}\\=\frac{1}{{{x^{2{\rm{ }} + {\rm{ }}3}}}}\\ = \underline{\underline {\frac{1}{{{x^5}}}}} \end{array}$$

$$\begin{array}{1}b)\qquad{(\frac{{{x^{ - 2}}}}{x})^{ - 2}}\\ = {({x^{ - 2{\rm{ }} - {\rm{ }}1}})^{ - 2}}\\= {({x^{ - 3}})^{ - 2}}\\ = {x^{ - 3{\rm{ }} \times {\rm{ }}( - 2)}}\\ = \underline{\underline {{x^6}}} \end{array}$$

$$\begin{array}{1}c)\qquad \frac{{{{({x^{ - 1}})}^0}}}{{({x^{ - 2}})({x^4})}}\\= \frac{1}{{{x^{ - 2{\rm{ }} + {\rm{ }}4}}}}\\ = \underline{\underline {\frac{1}{{{x^2}}}}} \end{array}$$

$$\begin{array}{l}d)\qquad {({x^{ - 1}}y)^2}{(2x{y^{ - 1}})^{ - 2}}\\ \\= ({x^{ - 1{\rm{ }} \times {\rm{ }}2}} \times {y^2}) \times [{2^{ - 2}} \times {x^{ - 2}} \times {y^{ - 1{\rm{ }} \times {\rm{ }}( - 2)}}]\\ = {x^{ - 2}} \times {y^2} \times {2^{ - 2}} \times {x^{ - 2}} \times {y^2}\\ = {2^{ - 2}} \times {x^{ - 2{\rm{ }} + {\rm{ }}( - 2)}} \times {y^{2{\rm{ }} + {\rm{ }}2}}\\ = {2^{ - 2}} \times {x^{ - 4}} \times {y^4}\\ = \underline{\underline {\frac{{{y^4}}}{{4{x^4}}}}} \end{array}$$

#### Understand the laws of rational indices

• For any real numbers a, b, and rational numbers p and q, the following arithmetic operations apply:

1. $${{\rm{a}}^{\rm{p}}}{a^q} = {a^{p + q}}$$

2. $$\frac{{{{\rm{a}}^{\rm{p}}}}}{{{a^q}}} = {a^{p - q}}$$

3. $${\left( {{{\rm{a}}^{\rm{p}}}} \right)^{\rm{q}}} = {a^{pq}}$$

4. $${{\rm{a}}^{\rm{p}}}{b^p} = {\left( {ab} \right)^p}$$

5. $$\frac{{{{\rm{a}}^{\rm{p}}}}}{{{b^p}}} = {\left( {\frac{a}{b}} \right)^p}$$

• If x and n are integers, simplify the following expressions.

$$a)\qquad {2^x} \times {4^x} \times {8^x}$$

$$b)\qquad \frac{{{5^x} \times {{({{25}^x})}^2}}}{{{{125}^x}}}$$

$$c)\qquad \frac{{({3^{x{\rm{ }} + {\rm{ }}1}})({9^{x{\rm{ }} - {\rm{ }}1}})}}{{{{({3^{2x{\rm{ }} + {\rm{ }}1}})}^2}}}$$

$$d)\qquad \frac{{{3^{4n{\rm{ }} - {\rm{ }}1}} + {9^{2n{\rm{ }} - {\rm{ }}3}}}}{{{{81}^{n{\rm{ }} - {\rm{ }}2}}}}$$

• $$\begin{array}{1}a)\qquad {2^x} \times {4^x} \times {8^x} \\= {2^x} \times {({2^2})^x} \times {({2^3})^x}\\ = {2^x} \times {2^{2x}} \times {2^{3x}}\\ = {2^{x{\rm{ }} + {\rm{ }}2x{\rm{ }} + {\rm{ }}3x}}\\ = \underline{\underline {{2^{6x}}}} \end{array}$$

$$\begin{array}{1}b)\qquad \frac{{{5^x} \times {{({{25}^x})}^2}}}{{{{125}^x}}} \\= \frac{{{5^x} \times {{({5^2})}^{2x}}}}{{{{({5^3})}^x}}}\\ = \frac{{{5^x} \times {5^{4x}}}}{{{5^{3x}}}}\\ = {5^{x{\rm{ }} + {\rm{ }}4x{\rm{ }} - {\rm{ }}3x}}\\ = \underline{\underline {{5^{2x}}}} \end{array}$$

$$\begin{array}{1}c)\qquad \frac{{({3^{x{\rm{ }} + {\rm{ }}1}})({9^{x{\rm{ }} - {\rm{ }}1}})}}{{{{({3^{2x{\rm{ }} + {\rm{ }}1}})}^2}}}\\ = \frac{{{3^{x{\rm{ }} + {\rm{ }}1}} \times {{({3^2})}^{x{\rm{ }} - {\rm{ }}1}}}}{{{3^{2(2x{\rm{ }} + {\rm{ }}1)}}}}\\ = \frac{{{3^{x{\rm{ }} + {\rm{ }}1}} \times {3^{2x{\rm{ }} - {\rm{ }}2}}}}{{{3^{4x{\rm{ }} + {\rm{ }}2}}}}\\ = {3^{x{\rm{ }} + {\rm{ }}1{\rm{ }} + {\rm{ }}2x{\rm{ }} - {\rm{ }}2{\rm{ }} - {\rm{ }}4x{\rm{ }} - {\rm{ }}2}}\\ = {3^{ - {\rm{ }}x{\rm{ }} - {\rm{ }}3}}\\ = {({3^{x{\rm{ }} + {\rm{ }}3}})^{ - 1}}\\ = \underline{\underline {\frac{1}{{{3^{x{\rm{ }} + {\rm{ }}3}}}}}} \end{array}$$

$$\begin{array}{1}d)\qquad\frac{{{3^{4n - 1}} + {9^{2n - 3}}}}{{{{81}^{n - 2}}}}\\ = \frac{{{3^{4n - 1}} + {{({3^2})}^{2n - 3}}}}{{{{({3^4})}^{n - 2}}}}\\ = \frac{{{3^{4n - 1}} + {3^{4n - 6}}}}{{{3^{4n - 8}}}}\\ = \frac{{{3^{4n}} \times {3^{ - 1}} + {3^{4n}} \times {3^{ - 6}}}}{{{3^{4n}} \times {3^{ - 8}}}}\\ = \frac{{({\textstyle{1 \over 3}} + {\textstyle{1 \over {{3^6}}}}){3^{4n}}}}{{{\textstyle{1 \over {{3^8}}}} \times {3^{4n}}}}\\ = (\frac{1}{3} + \frac{1}{{{3^6}}}) \times {3^8}\\ = \underline{\underline {2{\rm{ }}196}} \end{array}$$

#### Understand the definition and properties of logarithms (including the change of base)

• Definition of logarithms

If $${{\rm{a}}^{\rm{x}}} = N$$

Then $${\log _{\rm{a}}}N = x$$

Properties of logarithms

1. $${\log _{\rm{a}}}1 = 0$$

2. $${\log _{\rm{a}}}a = 1$$

3. $${\log _{\rm{a}}}MN = {\log _a}M + {\log _a}N$$

4. $${\log _{\rm{a}}}\frac{M}{N} = {\log _{\rm{a}}}M - {\log _a}N$$

5. $${\log _{\rm{b}}}N = \frac{{{{\log }_{\rm{a}}}N}}{{{{\log }_{\rm{a}}}b}}$$

• Find the values of the following expressions without using a calculator.

a) $$\frac{1}{5}\log 32 + \frac{1}{4}\log 25 - \log \sqrt 2$$

b) $$\frac{{\log 64}}{{\log 128}}$$

c) $$\frac{{\log \sqrt {{\textstyle{1 \over 3}}} }}{{\log 81}}$$

d) $${\log _2}8 - {\log _4}8$$

• $$\begin{array}{l}a)\qquad \frac{1}{5}\log 32 + \frac{1}{4}\log 25 - \log \sqrt 2 \\ = \log {({2^5})^{\frac{1}{5}}} + \log {({5^2})^{\frac{1}{4}}} - \log {2^{\frac{1}{2}}}\\ = \log 2 + \log {5^{\frac{1}{2}}} - \log {2^{\frac{1}{2}}}\\ = \log \frac{{2 \times {5^{\frac{1}{2}}}}}{{{2^{\frac{1}{2}}}}}\\ = \log ({2^{\frac{1}{2}}} \times {5^{\frac{1}{2}}})\\ = \log {10^{\frac{1}{2}}}\\ = \frac{1}{2}\log 10\\ = \underline{\underline {\frac{1}{2}}} \end{array}$$

$$\begin{array}{1}b)\qquad \frac{{\log 64}}{{\log 128}} \\= \frac{{\log {2^6}}}{{\log {2^7}}}\\ = \frac{{6\log 2}}{{7\log 2}}\\ = \underline{\underline {\frac{6}{7}}} \end{array}$$

$$\begin{array}{1}c)\qquad \frac{{\log \sqrt {{\textstyle{1 \over 3}}} }}{{\log 81}}\\ = \frac{{\log {{({3^{ - 1}})}^{\frac{1}{2}}}}}{{\log {3^4}}}\\ = \frac{{ - {\rm{ }}{\textstyle{1 \over 2}}\log 3}}{{4\log 3}}\\ = \underline{\underline { - {\rm{ }}\frac{1}{8}}} \end{array}$$

$$d)\qquad$$ $${\log _2}8 - {\log _4}8$$
$$= {\log _2}8 - \frac{{{{\log }_2}8}}{{{{\log }_2}4}}$$
$$= {\log _2}8 - \frac{{{{\log }_2}8}}{2}$$
$$\begin{array}{1} = \frac{1}{2}{\log _2}8 \\ =\underline{\underline {\frac{3}{2}}} \end{array}$$