Question

Simplify and write each of the following in exponential form:
(i) (25)³ ÷ 5³
(ii) (81)⁵ ÷ (3²)⁵
(iii) $\frac{9^8\times (x^2{)}^5}{(27{)}^4\times (x^3{)}^2}$
(iv) $\frac{3^2\times 7^8\times {13}^6}{{21}^2\times {91}^3}$

Answer 1

We have
(i) (25)³ ÷ 5³
= (5²)³÷ 5³
= 5⁶ ÷ 5³
= $\frac{5^6}{5^3}=5^{6-3}=5^3$

(ii) (81)⁵ ÷ (3²)⁵
= (3⁴)⁵ ÷ (3²)⁵
= (3)²⁰ ÷ (3)¹⁰
= $\frac{3^{20}}{3^{10}}=3^{20-10}=3^{10}$

(iii)

$\frac{9^8\times (x^2{)}^5}{(27{)}^4\times (x^3{)}^2}$
$=\frac{(3^2{)}^8\times (x^2{)}^5}{(3^3{)}^4\times (x^3{)}^2}$
$=\frac{3^{16}\times (x{)}^{10}}{3^{12}\times (x{)}^6}$
$=3$^16-12× (x)^10-6 = 3⁴× x⁴= (3x)⁴

(iv)

$\frac{3^2\times 7^8\times {13}^6}{{21}^2\times {91}^3}$
$=\frac{3^2\times 7^2\times 7^6\times {13}^6}{{21}^2\times (13\times 7{)}^3}$
$=\frac{(21{)}^2\times 7^6\times {13}^6}{{21}^2\times {13}^3\times 7^3}$
$=\frac{7^6\times {13}^6}{{13}^3\times 7^3}$
$=\frac{{91}^6}{{91}^3}={91}^{6-3}={91}^3$