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

$x=0$