Question

Question 15

Which of the following is not equal to 1?
(a)$\frac{2^3\times 3^2}{4\times 18}$
(b)$\left[\right(-2{)}^3\times (-2{)}^4]÷(-2{)}^7$
(c)$\frac{3^0\times 5^3}{5\times 25}$
(d)$\frac{2^4}{(7^0+3^0{)}^3}$

Answer 1

Let us solve the expressions,

For option (a),
$\frac{2^3\times 3^2}{4\times 18}=\frac{2\times 2\times 2\times 3\times 3}{4\times 18}=\frac{4\times 18}{4\times 18}=1$

For option (b),
$\left[\right(-2{)}^3\times (-2{)}^4]÷(-2{)}^7=\frac{(-2{)}^3\times (-2{)}^4}{(-2{)}^7}[\because a^m\times a^n=a^{m+n}]=\frac{(-2{)}^7}{(-2{)}^7}=1(\because \frac{a^m}{a^n}=a^{m-n})$

For option (c),
$\frac{3^0\times 5^3}{5\times 25}=\frac{1\times 5\times 5\times 5}{5\times 25}=\frac{5\times 25}{5\times 25}[\because a^0=1]=1$

For option (d),
$\frac{2^4}{(7^0+3^0{)}^3}=\frac{2^4}{(1+1{)}^3}[\because a^0=1]=\frac{2^4}{2^3}=2^{4-3}[\because \frac{a^m}{a^n}=a^{m-n},m>n]=2^1=2$

Hence, option (d) is not equal to 1.