To solve problems related to indices (or exponentials or powers), we need to have the list of laws of indices. So here are the laws/rules of indices.
• $a^0=1$
• $a^{-1}=1/a$
• $a^{-n}=1/a^n$
• $a^{m+n}=a^m . a^n$
• $(a^m)^n=a^{m \times n}$
• $\frac{a^m}{a^n}=a^{m-n}$
Solved Problems
Problem 1: Simplify $(81)^{3/4}$
Solution:
Note that $81=$ $3 \times 3 \times 3 \times 3$ $=3^4$
∴ $(81)^{3/4}$ $=(3^4)^{3/4}$
$=3^{4 \times 3/4}$ $[\because (a^m)^n=a^{m \times n}]$
$=3^3=27$
Problem 2: Find the value of $(8)^{-1/3}$ (cube root of 1/8)
Solution:
We have $8=$ $2 \times 2 \times 2$ $=2^3$
∴ $(8)^{-1/3}$ $=(2^3)^{-1/3}$
$=2^{-3 \times 1/3}$ $[\because (a^m)^n=a^{m \times n}]$
$=2^{}-1$
$=\frac{1}{2}$ $[\because a^{-1}=\frac{1}{a}]$
Problem 3: Simplify $(125)^{-1/3}$ (cube root of 1/125)
Solution:
Note that $125=$ $5 \times 5 \times 5$ $=5^3$
∴ $(125)^{-1/3}$ $=(5^3)^{-1/3}$
$=5^{-3 \times 1/3}$ $[\because (a^m)^n=a^{m \times n}]$
$=5^{-1}$
$=\frac{1}{5}$ $[\because a^{-1}=\frac{1}{a}]$
Problem 4: Find $\, 5^0 \times (16)^{-3/4}$
Solution:
Using $16=2^4,$ we get
$5^0 \times (16)^{-3/4}$
$=1 \times (2^4)^{-3/4}$ $[\because a^0=1]$
$=(2^4)^{-3/4}$
$=2^{4 \times \frac{-3}{4}}$ $[\because (a^m)^n=a^{m \times n}]$
$=2^{-3}$
$=\frac{1}{2^3}$ $[\because a^{-n}=\frac{1}{a^n}]$
$=\frac{1}{8}$
Problem 5: Simplify $\, x^{a-b}.x^{b-c}.x^{c-a}$
Solution:
$x^{a-b}.x^{b-c}.x^{c-a}$
$=x^{a-b+b-c+c-a}$
$=x^0$
$=1$
Problem 5: Solve for $x$
(i) $4^x=8^2$
(ii) $3^x=2^{-x}$
Solution:
(i) $4^x=8^2$
$\Rightarrow (2^2)^x=(2^3)^2$
$\Rightarrow 2^{2x}=2^{3 \times 2}$ $[\because (a^m)^n=a^{m \times n}]$
$\Rightarrow 2^{2x}=2^6$
Comparing the powers on both sides, we get
$2x=6$
$\Rightarrow x=\frac{6}{2}=3$
(ii) $3^x=2^{-x}$
$\Rightarrow 3^x=\frac{1}{2^x}$ $[\because a^{-n}=\frac{1}{a^n}]$
$\Rightarrow 3^x \times 2^x=1$
$\Rightarrow (3 \times 2)^x=1$ $[\because a^n \times b^n=(ab)^n]$
$\Rightarrow 6^x=6^0$
Comparing the powers, we obtain
$x=0$
