Indices And Logarithms

Indices and Logarithms

Indices (Law of Exponents)

Basic Rules

Three basic rules including the indices are:

i) $a^m \times a^n = a^{m+n}$

ii) $a^m \div a^n = a^{m-n}$

iii) $(a^m)^n = a^{mn}$

Negative Indices

Consider $a^5 \div a^2 = a^{5-2} = a^3$

$a^2 \div a^5 = a^{2-5} = a^{-3}$

In general

$a^{-m} = \dfrac{1}{a^m}$

Fractional Indices

Consider $a^{\frac{1}{2}} \times a^{\frac{1}{2}} = a^1$

$\rightarrow a^{\frac{1}{2}} = \sqrt{a}$

Similarly

$a^{\frac{1}{3}} \times a^{\frac{1}{3}} \times a^{\frac{1}{3}} = a^{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = a^1$

$\rightarrow a^{\frac{1}{3}} = \sqrt[3]{a}$

in general $a^{\frac{m}{n}} = \sqrt[n]{a^m}$

Zero Exponents

Consider $a^m \times a^0 = a^{m+0} \rightarrow a^0 = 1$

Logarithms

If a and b are two positive numbers there exist a third number c such that

$a^c = b$

$\rightarrow c$ is the logarithm of b to base a

i.e. $\log_a b = c$

Definition

Logarithm of 'x' to base 'a' is the power to which 'a' must be raised to give 'x'.

If $p = \log_a x$ and $q = \log_a y$, then $a^p = x$ and $a^q = y$

Thus:

$$\begin{aligned} xy &= a^{p+q} \\ \log_a(xy) &= \log_a(a^{p+q}) = p + q \\ &= \log_a x + \log_a y \\ \Rightarrow \log_a(xy) &= \log_a x + \log_a y \end{aligned}$$ $$\begin{aligned} \frac{x}{y} &= a^{p - q} \\ \log_a\left(\frac{x}{y}\right) &= \log_a(a^{p - q}) = p - q \\ &= \log_a x - \log_a y \\ \Rightarrow \log_a\left(\frac{x}{y}\right) &= \log_a x - \log_a y \end{aligned}$$

$$\begin{aligned} x^n &= a^{np} \\ \log_a(x^n) &= \log_a(a^{np}) = np \\ &= n \log_a x \\ \Rightarrow \log_a(x^n) &= n \log_a x \end{aligned}$$

Change of Base

If $y = \log_b a$, then $b^y = a$, taking $\log_c$ on both sides:

$$\frac{\log_c a}{\log_c b} = y$$ $$\log_b a = \frac{\log_c a}{\log_c b}$$ $$\log_b a = \frac{1}{\log_a b} \quad \text{if } c = a$$

Example

1. Solve for x, $\log_5 x + 2 \log_x 5 = 3$

Solution

$$\begin{aligned} \log_5 x + 2 \log_x 5 &= 3 \\ \log_5 x + \log_x 25 &= 3 \\ \log_5 x + \frac{1}{\log_{25} x} &= 3 \\ \log_5 x + \frac{1}{\frac{\log x}{\log 25}} &= 3 \\ \log_5 x + \frac{1}{\frac{\log x}{2 \log 5}} &= 3 \\ \log_5 x + \frac{2}{\log_5 x} &= 3 \\ \end{aligned}$$ $$y = \log_5 x$$ $$\begin{aligned} \log_{5}x + \frac{2}{\log_{5}x} &= 3 \\ y + \frac{2}{y} &= 3 \quad \text{(Let } y = \log_{5}x\text{)} \\ y^2 + 2 &= 3y \\ y^2 - 3y + 2 &= 0 \\ y^2 - y - 2y + 2 &= 0 \\ y(y - 1) - 2(y - 1) &= 0 \\ (y - 1)(y - 2) &= 0 \\ y &= 1 \text{ or } y = 2 \\ \log_{5}x &= 1 \text{ or } \log_{5}x = 2 \\ x &= 5 \text{ or } x = 25 \end{aligned}$$

Note that:

There are two important bases of logarithms

10 and e