Example 2

Find the value of the following.
(a) ${\log }_{2}7+{\log }_{2}12-{\log }_{2}21$

(b) $3{\log }_{10}5+2{\log }_{10}2-{\log }_{10}5$

(c) $2{\log }_{10}3-{\log }_{10}3+{\log }_{10}3\frac{1}{3}$

(d) ${\log }_{3}3p+{\log }_{3}3q-{\log }_{3}pq$

Example 1

Express the following in term of ${\log }_{a}x$   and ${\log }_{a}y$  .
(a) ${\log }_{a}3x$
(b) ${\log }_a\frac{x}{5}$
(c) ${\log }_a{y}^5$
(d) ${\log }_{a}x{y}^3$
(e) ${\log }_a\frac{x^2}{\sqrt{y}}$
(f) ${\log }_a\sqrt{\frac{y}{a^2{x}^3}}$

Law of Logarithms

$$\boxed{\begin{array}{l}\text{ Law 1:}\\ {\text{ log}}_{a}xy={\text{log}}_{a}x+{\text{log}}_{a}y\\ \\ \text{ Example:}\\ {\log }_{5}25x={\text{log}}_{5}25+{\text{log}}_{5}x\\ \\ \text{ Beware!!}\\ {\text{ log}}_{a}x+{\text{log}}_{a}y\ne {\text{log}}_a\left(x+y\right)\end{array}}$$

$$\boxed{\begin{array}{l}\text{ Law 2:}\\ {\text{ log}}_a\left(\frac{x}{y}\right)={\text{log}}_{a}x-{\text{log}}_{a}y\\ \\ \text{ Example:}\\ {\log }_5\frac{x}{25}={\text{log}}_{5}x-{\text{log}}_{5}25\\ \\ \text{ Beware!!}\\ {\text{ log}}_a\frac{x}{y}\ne \frac{{\text{log}}_{a}x}{{\text{log}}_{a}y}\end{array}}$$

$$\boxed{\begin{array}{l}\text{ Law 3:}\\ {\text{ log}}_a{x}^m=m{\text{log}}_{a}x\\ \\ \text{ Example:}\\ {\log }_5{y}^5=5{\text{log}}_{5}y\\ \\ \text{ Beware!!}\\ {\left({\text{log}}_{a}x\right)}^2\ne 2{\text{log}}_{a}x\end{array}}$$

Note:

1. The logarithm of a negative number is not defined.
2. log in the calculator denotes ${\log }_{10}$ or common logarithm.
3. ${\log }_{10}$ may be written as lg.
4. If the base is other than 10, the base should be specified, e.g. ${\log }_{3}81$

$\boxed{\begin{array}{l}{a}^{\frac{1}{n}}\text{ is a }n\text{th root of }a.\\ a^{\frac{1}{n}}=\sqrt[n]{a}\\ a^{\frac{m}{n}}\text{ is a }n\text{th root of }{a}^m.\\ a^{\frac{m}{n}}=\sqrt[n]{a^m}\end{array}}$

Find the value of the followings:
$\begin{array}{l}{\text{(a) 81}}^{\frac{1}{2}}\\ {\text{(b) 64}}^{\frac{1}{3}}\\ {\text{(c) 625}}^{\frac{1}{4}}\end{array}$

$\begin{array}{l}{\text{(a) 81}}^{\frac{1}{2}}=\sqrt{81}=9\\ {\text{(b) 64}}^{\frac{1}{3}}=\sqrt[3]{64}=4\\ {\text{(c) 625}}^{\frac{1}{4}}=\sqrt[4]{625}=5\end{array}$

$\begin{array}{l}\text{(a) }{16}^{\frac{3}{2}}\\ \text{(b) }{\left(\frac{27}{64}\right)}^{\frac{2}{3}}\end{array}$

Solution:
$\begin{array}{l}\text{(a) }{16}^{\frac{3}{2}}={\left({16}^{\frac{1}{2}}\right)}^3=4^3=64\\ \text{(b) }{\left(\frac{27}{64}\right)}^{\frac{2}{3}}={\left(\sqrt[3]{\frac{27}{64}}\right)}^2={\left(\frac{3}{4}\right)}^2=\frac{9}{16}\end{array}$

$\boxed{\begin{array}{l}{a}^m\times a^n=a^{m+n}\\ \\ \text{ Example:}\\ 3^3\times 3^2\\ =3^{3+2}=3^5=243\end{array}}$


$\boxed{\begin{array}{l}{a}^m÷a^n=a^{m-n}\text{ or }\frac{a^m}{a^n}=a^{m-n},a\ne 0\\ \\ \text{ Example:}\\ 3^3÷3^2\\ =3^{3-2}=3^1=3\\ \text{ or}\\ \frac{3^3}{3^2}=3^{3-2}=3^1=3\end{array}}$


$\boxed{\begin{array}{l}{\left(a^m\right)}^n=a^{mn}\\ \\ \text{ Example:}\\ {\left(7^3\right)}^4=7^{3\times 4}=7^{12}\end{array}}$


$\boxed{\begin{array}{l}{\left(ab\right)}^n=a^n{b}^n\\ \\ \text{ Example:}\\ {\left(15\right)}^3={\left(5\times 3\right)}^3=2^3\times 3^3\\ \end{array}}$


$\boxed{\begin{array}{l}{\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n},b\ne 0\\ \\ \text{ Example:}\\ {\left(\frac{3}{5}\right)}^4=\frac{3^4}{5^4}=\frac{81}{625}\end{array}}$

a 0 = 1, where a ≠ 0

$\begin{array}{l}\text{(c) }{\left(\frac{2}{7}\right)}^0=1\\ \text{(d) }{\left(-11\frac{1}{25}\right)}^0=1\end{array}$

$\begin{array}{l}\text{(c) }{\left(\frac{1}{3}\right)}^{-4}\\ \text{(d) }{\left(\frac{2}{5}\right)}^{-2}\\ \text{(e) }-{\left(\frac{2}{5}\right)}^{-4}\end{array}$

Solution:
$\begin{array}{l}\text{(a) }{102}^{-1}=\frac{1}{102}\\ \text{(b) }-6^{-3}=\frac{1}{-6^3}=-\frac{1}{216}\\ \text{(c) }{\left(\frac{1}{3}\right)}^{-4}={\left(3\right)}^4=81\\ \text{(d) }{\left(\frac{2}{5}\right)}^{-2}={\left(\frac{5}{2}\right)}^2=\frac{25}{4}\\ \text{(e) }{\left(-\frac{2}{5}\right)}^{-4}={\left(-\frac{5}{2}\right)}^4=\frac{625}{16}\end{array}$