Exponents — Exercise 5(A)

Find the value of :

(i) 6²

(ii) 7³

(iii) 4⁴

(iv) 5⁵

(v) 8³

(vi) 7⁵

Answer

(i) Solving,

⇒ 6² = 6 × 6 = 36.

Hence, 6² = 36.

(ii) Solving,

⇒ 7³ = 7 × 7 × 7 = 343.

Hence, 7³ = 343.

(iii) Solving,

⇒ 4⁴ = 4 × 4 × 4 × 4 = 256.

Hence, 4⁴ = 256.

(iv) Solving,

⇒ 5⁵ = 5 × 5 × 5 × 5 × 5 = 3125.

Hence, 5⁵ = 3125.

(v) Solving,

⇒ 8³ = 8 × 8 × 8 = 512.

Hence, 8³ = 512.

(vi) Solving,

⇒ 7⁵ = 7 × 7 × 7 × 7 × 7 = 16807.

Hence, 7⁵ = 16807.

Evaluate :

(i) 2³ × 4²

(ii) 2³ × 5²

(iii) 3³ × 5²

(iv) 2² × 3³

(v) 3² × 5³

(vi) 5³ × 2⁴

(vii) 3² × 4²

(viii) (4 × 3)³

(ix) (5 × 4)²

Answer

(i) Solving,

⇒ 2³ × 4² = (2 × 2 × 2) × (4 × 4) = 8 × 16 = 128.

Hence, 2³ × 4² = 128.

(ii) Solving,

⇒ 2³ × 5² = (2 × 2 × 2) × (5 × 5) = 8 × 25 = 200.

Hence, 2³ × 5² = 200.

(iii) Solving,

⇒ 3³ × 5² = (3 × 3 × 3) × (5 × 5) = 27 × 25 = 675.

Hence, 3³ × 5² = 675.

(iv) Solving,

⇒ 2² × 3³ = (2 × 2) × (3 × 3 × 3) = 4 × 27 = 108.

Hence, 2² × 3³ = 108.

(v) Solving,

⇒ 3² × 5³ = (3 × 3) × (5 × 5 × 5) = 9 × 125 = 1125.

Hence, 3² × 5³ = 1125.

(vi) Solving,

⇒ 5³ × 2⁴ = (5 × 5 × 5) × (2 × 2 × 2 × 2) = 125 × 16 = 2000.

Hence, 5³ × 2⁴ = 2000.

(vii) Solving,

⇒ 3² × 4² = (3 × 3) × (4 × 4) = 9 × 16 = 144.

Hence, 3² × 4² = 144.

(viii) Solving,

⇒ (4 × 3)³ = 12³ = 12 × 12 × 12 = 1728.

Hence, (4 × 3)³ = 1728.

(ix) Solving,

⇒ (5 × 4)² = 20² = 20 × 20 = 400.

Hence, (5 × 4)² = 400.

Evaluate :

$\left(\dfrac{3}{4}\right)^4$

Answer

Solving,

$\Rightarrow \left(\dfrac{3}{4}\right)^4 \\[1em] = \dfrac{3 \times 3 \times 3 \times 3}{4 \times 4 \times 4 \times 4} \\[1em] = \dfrac{81}{256}$

Hence, $\left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256}$.

Evaluate :

$\left(-\dfrac{5}{6}\right)^5$

Answer

Solving,

$\Rightarrow \left(-\dfrac{5}{6}\right)^5 \\[1em] = \dfrac{(-5) \times (-5) \times (-5) \times (-5) \times (-5)}{6 \times 6 \times 6 \times 6 \times 6} \\[1em] = -\dfrac{3125}{7776}$

Hence, $\left(-\dfrac{5}{6}\right)^5 = -\dfrac{3125}{7776}$.

Evaluate :

$\left(\dfrac{-3}{-5}\right)^3$

Answer

Solving,

$\Rightarrow \left(\dfrac{-3}{-5}\right)^3\\[1em] = \left(\dfrac{3}{5}\right)^3 \\[1em] = \dfrac{3 \times 3 \times 3}{5 \times 5 \times 5} \\[1em] = \dfrac{27}{125}$

Hence, $\left(\dfrac{-3}{-5}\right)^3 = \dfrac{27}{125}$.

Evaluate :

$\left(\dfrac{2}{3}\right)^3 \times \left(\dfrac{3}{4}\right)^2$

Answer

Solving,

$\Rightarrow \left(\dfrac{2}{3}\right)^3 \times \left(\dfrac{3}{4}\right)^2\\[1em] = \dfrac{2 \times 2 \times 2}{3 \times 3 \times 3} \times \dfrac{3 \times 3}{4 \times 4}\\[1em] = \dfrac{2 \times 2 \times 2 \times 3 \times 3}{3 \times 3 \times 3 \times 4 \times 4}\\[1em] = \dfrac{8}{3 \times 16}\\[1em] = \dfrac{8}{48}\\[1em] = \dfrac{1}{6}$

Hence, $\left(\dfrac{2}{3}\right)^3 \times \left(\dfrac{3}{4}\right)^2 = \dfrac{1}{6}$.

Evaluate :

$\left(-\dfrac{3}{4}\right)^3 \times \left(\dfrac{2}{3}\right)^4$

Answer

Solving,

$\Rightarrow \left(-\dfrac{3}{4}\right)^3 \times \left(\dfrac{2}{3}\right)^4\\[1em] = \dfrac{(-3) \times (-3) \times (-3)}{4 \times 4 \times 4} \times \dfrac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3}\\[1em] = -\dfrac{27}{64} \times \dfrac{16}{81}\\[1em] = -\dfrac{27 \times 16}{64 \times 81}\\[1em] = -\dfrac{1}{4 \times 3}\\[1em] = -\dfrac{1}{12}$

Hence, $\left(-\dfrac{3}{4}\right)^3 \times \left(\dfrac{2}{3}\right)^4 = -\dfrac{1}{12}$.

Evaluate :

$\left(\dfrac{3}{5}\right)^2 \times \left(-\dfrac{2}{3}\right)^3$

Answer

Solving,

$\Rightarrow \left(\dfrac{3}{5}\right)^2 \times \left(-\dfrac{2}{3}\right)^3\\[1em] = \dfrac{3 \times 3}{5 \times 5} \times \dfrac{(-2) \times (-2) \times (-2)}{3 \times 3 \times 3}\\[1em] = \dfrac{9}{25} \times \left(-\dfrac{8}{27}\right)\\[1em] = -\dfrac{9 \times 8}{25 \times 27}\\[1em] = -\dfrac{8}{25 \times 3}\\[1em] = -\dfrac{8}{75}$

Hence, $\left(\dfrac{3}{5}\right)^2 \times \left(-\dfrac{2}{3}\right)^3 = -\dfrac{8}{75}$.

Which is greater :

(i) 2³ or 3²

(ii) 2⁵ or 5²

(iii) 4³ or 3⁴

(iv) 5⁴ or 4⁵

Answer

(i) Solving,

⇒ 2³ = 2 × 2 × 2 = 8

⇒ 3² = 3 × 3 = 9.

Since 9 > 8, therefore 3² > 2³.

Hence, the greater number is 3².

(ii) Solving,

⇒ 2⁵ = 2 × 2 × 2 × 2 × 2 = 32

⇒ 5² = 5 × 5 = 25.

Since 32 > 25, therefore 2⁵ > 5².

Hence, the greater number is 2⁵.

(iii) Solving,

⇒ 4³ = 4 × 4 × 4 = 64

⇒ 3⁴ = 3 × 3 × 3 × 3 = 81.

Since 81 > 64, therefore 3⁴ > 4³.

Hence, the greater number is 3⁴.

(iv) Solving,

⇒ 5⁴ = 5 × 5 × 5 × 5 = 625

⇒ 4⁵ = 4 × 4 × 4 × 4 × 4 = 1024.

Since 1024 > 625, therefore 4⁵ > 5⁴.

Hence, the greater number is 4⁵.

Express 512 in exponential form.

Answer

By prime factorisation of 512:

$\begin{array}{l|r} 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

⇒ 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹.

Hence, 512 = 2⁹.

Express 1250 in exponential form.

Answer

By prime factorisation of 1250:

$\begin{array}{l|r} 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

⇒ 1250 = 2 × 5 × 5 × 5 × 5 = 2¹ × 5⁴.

Hence, 1250 = 2 × 5⁴.

Express 1458 in exponential form.

Answer

By prime factorisation of 1458:

$\begin{array}{l|r} 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

⇒ 1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3 = 2¹ × 3⁶.

Hence, 1458 = 2 × 3⁶.

Express 3600 in exponential form.

Answer

By prime factorisation of 3600:

$\begin{array}{l|r} 2 & 3600 \\ \hline 2 & 1800 \\ \hline 2 & 900 \\ \hline 2 & 450 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

⇒ 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴ × 3² × 5².

Hence, 3600 = 2⁴ × 3² × 5².

Express 1350 in exponential form.

Answer

By prime factorisation of 1350:

$\begin{array}{l|r} 2 & 1350 \\ \hline 3 & 675 \\ \hline 3 & 225 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

⇒ 1350 = 2 × 3 × 3 × 3 × 5 × 5 = 2¹ × 3³ × 5².

Hence, 1350 = 2 × 3³ × 5².

Express 1176 in exponential form.

Answer

By prime factorisation of 1176:

$\begin{array}{l|r} 2 & 1176 \\ \hline 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

⇒ 1176 = 2 × 2 × 2 × 3 × 7 × 7 = 2³ × 3¹ × 7².

Hence, 1176 = 2³ × 3 × 7².

If a = 2 and b = 3, find the value of :

(i) (a + b)²

(ii) (b - a)³

(iii) (a × b)^a

(iv) (a × b)^b

Answer

(i) Solving,

⇒ (a + b)² = (2 + 3)² = 5² = 5 × 5 = 25.

Hence, (a + b)² = 25.

(ii) Solving,

⇒ (b - a)³ = (3 - 2)³ = 1³ = 1 × 1 × 1 = 1.

Hence, (b - a)³ = 1.

(iii) Solving,

⇒ (a × b)^a = (2 × 3)² = 6² = 6 × 6 = 36.

Hence, (a × b)^a = 36.

(iv) Solving,

⇒ (a × b)^b = (2 × 3)³ = 6³ = 6 × 6 × 6 = 216.

Hence, (a × b)^b = 216.

Express :

(i) 1024 as a power of 2.

(ii) 343 as a power of 7.

(iii) 729 as a power of 3.

Answer

(i) By prime factorisation of 1024:

$\begin{array}{l|r} 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

⇒ 1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2¹⁰.

Hence, 1024 = 2¹⁰.

(ii) By prime factorisation of 343:

$\begin{array}{l|r} 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

⇒ 343 = 7 × 7 × 7 = 7³.

Hence, 343 = 7³.

(iii) By prime factorisation of 729:

$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

⇒ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶.

Hence, 729 = 3⁶.

If 27 × 32 = 3^x × 2^y; find the values of x and y.

Answer

By prime factorisation of 27:

$\begin{array}{l|r} 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

⇒ 27 = 3 × 3 × 3 = 3³.

By prime factorisation of 32:

$\begin{array}{l|r} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

⇒ 32 = 2 × 2 × 2 × 2 × 2 = 2⁵.

So,

⇒ 27 × 32 = 3^x × 2^y

⇒ 3³ × 2⁵ = 3^x × 2^y.

Comparing the powers of the same base on both sides, we get :

⇒ x = 3 and y = 5.

Hence, x = 3 and y = 5.

If 64 × 625 = 2^a × 5^b; find :

(i) the values of a and b.

(ii) 2^b × 5^a

Answer

(i) Solving,

⇒ 64 × 625 = 2^a × 5^b

⇒ (2 × 2 × 2 × 2 × 2 × 2) × (5 × 5 × 5 × 5) = 2^a × 5^b

⇒ 2⁶ × 5⁴ = 2^a × 5^b.

Comparing the powers of the same base on both sides, we get :

⇒ a = 6 and b = 4.

Hence, a = 6 and b = 4.

(ii) Solving,

⇒ 2^b × 5^a = 2⁴ × 5⁶

⇒ (2 × 2 × 2 × 2) × (5 × 5 × 5 × 5 × 5 × 5)

⇒ 16 × 15625

⇒ 250000.

Hence, 2^b × 5^a = 250000.