Solved Problems on Indices

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

$x=0$