DAV Class 7 Maths Chapter 4 Worksheet 4

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DAV Class 7 Maths Chapter 4 Worksheet 4

Exponents and Powers Worksheet 4


1. Fill in the blanks.

(i) \( \displaystyle (16)^{15} \div (16)^{19} = \left(\frac{1}{16^?}\right) \)

Solution

\begin{align*} &= (16)^{15} \div (16)^{19} \\ &= (16)^{15-19} \\ &= (16)^{-4} \\ &= \frac{1}{16^4} \end{align*}

Anwer \( \displaystyle (16)^{15} \div (16)^{19} =\frac{1}{16^{\color{red} \boxed4}}\)

(ii) \( \displaystyle \left(\frac{11}{12}\right)^? \div \left(\frac{11}{12}\right)^{22} = \frac{1}{\left(\frac{11}{12}\right)^2} \)

Solution

\begin{align*} \left(\frac{11}{12}\right)^? \div \left(\frac{11}{12}\right)^{22} &= \left(\frac{11}{12}\right)^{-2} \\ \\ \left(\frac{11}{12}\right)^? &= \left(\frac{11}{12}\right)^{-2} \times \left(\frac{11}{12}\right)^{22} \\ \\ &= \left(\frac{11}{12}\right)^{-2 + 22} \\ \\ &= \left(\frac{11}{12}\right)^{20} \end{align*}

Anwer \( \displaystyle \left(\frac{11}{12}\right)^{\color{red}\boxed{20}} \div \left(\frac{11}{12}\right)^{22} = \frac{1}{\left(\frac{11}{12}\right)^2} \)

(iii) \( \displaystyle \left(\frac{1}{3^2}\right)^? = \left(\frac{1}{3}\right)^6 \)

Solution

\begin{align*} &= \left(\frac{1}{3}\right)^6 \\ \\ &= \left(\frac{1}{3}\right)^{2 \times 3} \\ \\ &= \left(\frac{1}{3^2}\right)^{3} \end{align*}

Anwer \( \displaystyle \left(\frac{1}{3^2}\right)^{\color{red}\boxed3} = \left(\frac{1}{3}\right)^6 \)

(iv) \( \displaystyle \left[\left(-\frac{1}{2}\right)^4\right]^2 = \left(-\frac{1}{2}\right)^{?} \)

Solution

\begin{align*} &= \left[\left(-\frac{1}{2}\right)^4\right]^2 \\ \\ &= \left(-\frac{1}{2}\right)^{4 \times 2} \\ \\ &= \left(-\frac{1}{2}\right)^8 \end{align*}

Anwer \( \displaystyle \left[\left(-\frac{1}{2}\right)^4\right]^2 = \left(-\frac{1}{2}\right)^{\color{red} \boxed{8}} \)

(v) \( \displaystyle \left(\frac{1}{12}\right)^5 \div \left(\frac{1}{12}\right)^5 = \left(\frac{1}{12}\right)^? \)

Solution

\begin{align*} & = \left(\frac{1}{12}\right)^5 \div \left(\frac{1}{12}\right)^5 \\ \\ &= \left(\frac{1}{12}\right)^{5-5} \\ \\ & = \left(\frac{1}{12}\right)^0 \end{align*}

Anwer \( \displaystyle \left(\frac{1}{12}\right)^5 \div \left(\frac{1}{12}\right)^5 = \left(\frac{1}{12}\right)^{\color{red}\boxed{0}} \)

(vi) \( \displaystyle \left[\left(-\frac{1}{7}\right)^0\right]^7 = \left(-\frac{1}{7}\right)^{?} \)

Solution

\begin{align*} &= \left(-\frac{1}{7}\right)^{0 \times 7} \\ \\ &= \left(-\frac{1}{7}\right)^0 \end{align*}

Anwer \( \displaystyle \left[\left(-\frac{1}{7}\right)^0\right]^7 = \left(-\frac{1}{7}\right)^{\color{red}\boxed{0}} \)

2. Simplify and express the result in exponential form.

(i) \( \displaystyle \left[\left(\frac{9}{2}\right)^0\right]^5 \)

Solution

\begin{align*} &= \left(\frac{9}{2}\right)^{0 \times 5} \\ \\ &= \left(\frac{9}{2}\right)^{0} \end{align*}

Anwer \( \displaystyle \left[\left(\frac{9}{2}\right)^0\right]^5 = \color{red} \left(\frac{9}{2}\right)^{0} \)

(ii) \( \displaystyle \left[\left(-\frac{5}{17}\right)^6\right]^3 \)

Solution

\begin{align*} &= \left(-\frac{5}{17}\right)^{6 \times 3} \\ \\ &= \left(-\frac{5}{17}\right)^{18} \end{align*}

Anwer \( \displaystyle \left[\left(-\frac{5}{17}\right)^6\right]^3 = \color{red} \left(-\frac{5}{17}\right)^{18}\)

(iii) \( \displaystyle \left[\left(\frac{4}{7}\right)^4\right]^4 \)

Solution

\begin{align*} &= \left(\frac{4}{7}\right)^{4 \times 4} \\ \\ &= \left(\frac{4}{7}\right)^{16} \end{align*}

Anwer \( \displaystyle \left[\left(\frac{4}{7}\right)^4\right]^4 = \color{red} \left(\frac{4}{7}\right)^{16} \)

(iv) \( \displaystyle \left[ \left(-\frac{2}{5}\right)^3 \times \left(-\frac{2}{5}\right)^2 \right]^4 \)

Solution

\begin{align*} & = \left[ \left(-\frac{2}{5}\right)^3 \times \left(-\frac{2}{5}\right)^2 \right]^4 \\ \\ & = \left[ \left(-\frac{2}{5}\right)^{3 + 2} \right]^4 \\ \\ & = \left[ \left(-\frac{2}{5}\right)^5 \right]^4 \\ \\ & = \left(-\frac{2}{5}\right)^{5 \times 4} \\ \\ & = \left(-\frac{2}{5}\right)^{20} \end{align*}

Anwer \( \displaystyle \left[ \left(-\frac{2}{5}\right)^3 \times \left(-\frac{2}{5}\right)^2 \right]^4 = \color{red} \left(-\frac{2}{5}\right)^{20} \)

(v) \( \displaystyle \left(\frac{3}{5}\right)^3 \div \left(\frac{3}{5}\right)^8 \)

Solution

\begin{align*} \color{magenta} a^m \div a^n &= \color{magenta} \frac{1}{a^{n- m}}, \ n > m \\ \\ & = \left(\frac{3}{5}\right)^3 \div \left(\frac{3}{5}\right)^8 \\ \\ & = \frac{1}{\left(\frac{3}{5}\right)^{8 - 3}} \\ \\ & = \frac{1}{\left(\frac{3}{5}\right)^{5}} \\ \\ & = \left(\frac{5}{3}\right)^{5} \end{align*}

Anwer \( \displaystyle \left(\frac{3}{5}\right)^3 \div \left(\frac{3}{5}\right)^8 = \color{red} \left(\frac{5}{3}\right)^{5} \)

(vi) \( \displaystyle \left[\left(\frac{12}{5}\right)^0 \times \left(\frac{12}{5}\right)\right]^5 \)

Solution

\begin{align*} & = \left[\left(\frac{12}{5}\right)^{0 + 1}\right]^5 \\ \\ & = \left[\left(\frac{12}{5}\right)^{1}\right]^5 \\ \\ & = \left(\frac{12}{5}\right)^{1 \times 5} \\ \\ & = \left(\frac{12}{5}\right)^{5} \end{align*}

Anwer \( \displaystyle \left[\left(\frac{12}{5}\right)^0 \times \left(\frac{12}{5}\right)\right]^5 = \color{red} \left(\frac{12}{5}\right)^{5} \)

3. Evaluate

(i) \( \displaystyle \left(-\frac{3}{4}\right)^3 \div \left(-\frac{3}{4}\right)^5 \)

Solution

\begin{align*} \color{magenta} a^m \div a^n &= \color{magenta} \frac{1}{a^{n- m}}, \ n > m \\ \\ &= \left(-\frac{3}{4}\right)^3 \div \left(-\frac{3}{4}\right)^5 \\ \\ & = \frac{1}{\left(-\frac{3}{4}\right)^{5 - 3}} \\ \\ & = \frac{1}{\left(-\frac{3}{4}\right)^{2}} \\ \\ & = {\left(-\frac{4}{3}\right)^{2}} \\ \\ & = \left(-\frac{4}{3}\right) \times \left(-\frac{4}{3}\right) \\ \\ & = \frac{16}{9} \end{align*}

Anwer \( \displaystyle \color{red} \frac{16}{9} \)

(ii) \( \displaystyle \left(\frac{1}{5^2}\right)^2 \times \frac{1}{5} \)

Solution

\begin{align*} & = \left(\frac{1}{5^2}\right)^2 \times \frac{1}{5} \\ \\ & = \left(\frac{1}{5^{2 \times 2}}\right) \times \frac{1}{5} \\ \\ & = \frac{1}{5^{4}} \times \frac{1}{5} \\ \\ & = \frac{1}{5^{4+1}} \\ \\ & = \frac{1}{5^{5}} \\ \\ & = \frac{1}{5 \times 5 \times 5 \times 5 \times 5} \\ \\ & = \frac{1}{3125} \end{align*}

Anwer \( \displaystyle \color{red} \frac{1}{3125} \)

(iii) \( \displaystyle \left[\left(-\frac{5}{6}\right)^2\right]^2 \div \left(-\frac{5}{6}\right)^2 \)

Solution

\begin{align*} &= \left(-\frac{5}{6}\right)^{2 \times 2} \div \left(-\frac{5}{6}\right)^2 \\ \\ &= \left(-\frac{5}{6}\right)^{4} \div \left(-\frac{5}{6}\right)^2 \\ \\ &= \left(-\frac{5}{6}\right)^{4-2} \\ \\ &= \left(-\frac{5}{6}\right)^{2} \\ \\ &= \left(-\frac{5}{6}\right) \times \left(-\frac{5}{6}\right) \\ \\ &= \frac{25}{36} \end{align*}

Anwer \( \displaystyle \color{red} \frac{25}{36} \)

(iv) \( \displaystyle \left(\frac{2}{3}\right)^2 \div \left[\left(\frac{2}{3}\right)^2\right]^0 \)

Solution

\begin{align*} &= \left(\frac{2}{3}\right)^2 \div \left(\frac{2}{3}\right)^{2 \times 0 } \\ \\ &= \left(\frac{2}{3}\right)^2 \div \left(\frac{2}{3}\right)^{0} \\ \\ &= \left(\frac{2}{3}\right)^{2 - 0} \\ \\ &= \left(\frac{2}{3}\right)^{2} \\ \\ &= \frac{2}{3} \times \frac{2}{3} \\ \\ &= \frac{4}{9} \end{align*}

Anwer \( \displaystyle \color{red} \frac{4}{9} \)

(v) \( \displaystyle \frac{\left(\frac{1}{2}\right)^5}{\left(\frac{1}{2}\right)^3} - \frac{\left(\frac{1}{2}\right)^6}{\left(\frac{1}{2}\right)^5} \)

Solution

\begin{align*} &= \left[\left(\frac{1}{2}\right)^{5} \div \left(\frac{1}{2}\right)^3\right] - \left[\left(\frac{1}{2}\right)^{6} \div \left(\frac{1}{2}\right)^5 \right] \\ \\ &= \left(\frac{1}{2}\right)^{5-3} - \left(\frac{1}{2}\right)^{6-5} \\ \\ &= \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^1 \\ \\ &= \frac{1}{4} - \frac{1}{2 } \\ \\ &= \frac{1}{4} - \frac{1 {\color{green}\times 2}}{2 {\color{green}\times 2}} \\ \\ &= \frac{1 - 2}{4}\\ \\ &= \frac{-1}{4} \end{align*}

Anwer \( \displaystyle \color{red}\frac{-1}{4} \)

4. Find the value of:

(i) \( \displaystyle \left(\frac{5}{7}\right)^{3 \times 2 - 6} \)

Solution

\begin{align*} & = \left(\frac{5}{7}\right)^{3 \times 2 - 6} \\ & = \left(\frac{5}{7}\right)^{6 - 6} \\ & = \left(\frac{5}{7}\right)^{0} \\ & = \color{red} 1 \end{align*}

(ii) \( \displaystyle (3^0 + 4^0) \times 5^0 \)

Solution

\begin{align*} &= (3^0 + 4^0) \times 5^0 \\ &= (1 + 1) \times 1 \\ &= 2 \times 1 \\ &= \color{red} 2 \\ \end{align*}

(iii) \( \displaystyle 1^0 \times 2^0 + 3^0 \times 4^0 + 5^0 \times 6^0 \)

Solution

\begin{align*} &= 1^0 \times 2^0 + 3^0 \times 4^0 + 5^0 \times 6^0 \\ &= {\color{green} 1 \times 1} + {\color{magenta} 1 \times 1 } + {\color{brown} 1 \times 1} \\ &= 1 + 1 + 1 \\ &= \color{red} 3 \\ \end{align*}

(iv) \( \displaystyle \left(-\frac{5}{9}\right)^{9 - 3 \times 2 - 3} \)

Solution

\begin{align*} &= \left(-\frac{5}{9}\right)^{9 - 3 \times 2 - 3} \\ \\ &= \left(-\frac{5}{9}\right)^{9 - 6 - 3} \\ \\ &= \left(-\frac{5}{9}\right)^{3 - 3} \\ \\ &= \left(-\frac{5}{9}\right)^{0} \\ \\ &= \color{red} 1 \end{align*}

5. Find the value of x so that -

(i) \( \displaystyle \left(\frac{3}{4}\right)^{2x + 1} = \left[\left(\frac{3}{4}\right)^3\right]^3 \)

Solution

\begin{align*} \left(\frac{3}{4}\right)^{2x + 1} &= \left(\frac{3}{4}\right)^{3 \times 3} \\ \\ \left(\frac{3}{4}\right)^{\color{magenta}2x + 1} &= \left(\frac{3}{4}\right)^{\color{magenta}9} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \color{magenta}2x + 1 &= \color{magenta}9 \\ 2x &= 9 - 1 \\ 2x &= 8 \\ x &= \frac{\cancel8^4}{\cancel2_1} \\ \color{red} x &= \color{red}4 \end{align*}

(ii) \( \displaystyle \left(\frac{2}{5}\right)^{3} \times \left(\frac{2}{5}\right)^{6} = \left(\frac{2}{5}\right)^{3x} \)

Solution

\begin{align*} \left(\frac{2}{5}\right)^{3 + 6} &= \left(\frac{2}{5}\right)^{3x} \\ \\ \left(\frac{2}{5}\right)^{\color{magenta}9} &= \left(\frac{2}{5}\right)^{\color{magenta}3x} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \color{magenta}9 &= \color{magenta}3x \\ x &= \frac{\cancel9^3}{\cancel3_1} \\ \color{red} x &= \color{red} 3 \end{align*}

(iii) \( \displaystyle \left(-\frac{1}{5}\right)^{20} \div {\left(-\frac{1}{5}\right)^{15}} = \left(-\frac{1}{5}\right)^{5x} \)

Solution

\begin{align*} \left(-\frac{1}{5}\right)^{20 - 15} &= \left(-\frac{1}{5}\right)^{5x} \\ \\ \left(-\frac{1}{5}\right)^{\color{magenta}5} &= \left(-\frac{1}{5}\right)^{\color{magenta}5x} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \color{magenta}5x &= \color{magenta}5 \\ x &= \frac{\cancel5^1}{\cancel5_1} \\ \color{red}x &= \color{red}1 \end{align*}

(iv) \( \displaystyle \left(\frac{1}{16}\right) \times \left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{3(x - 2)} \)

Solution

\begin{align*} \left(\frac{1}{16}\right) \times \left(\frac{1}{2}\right)^2 &= \left(\frac{1}{2}\right)^{3(x - 2)} \\ \\ \left(\frac{1}{2^4}\right) \times \left(\frac{1}{2}\right)^2 &= \left(\frac{1}{2}\right)^{3(x - 2)} \\ \\ \left(\frac{1}{2}\right)^4 \times \left(\frac{1}{2}\right)^2 &= \left(\frac{1}{2}\right)^{3(x - 2)} \\ \\ \left(\frac{1}{2}\right)^{4+2} &= \left(\frac{1}{2}\right)^{3(x - 2)} \\ \\ \left(\frac{1}{2}\right)^{\color{magenta}6} &= \left(\frac{1}{2}\right)^{\color{magenta}3(x - 2)} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \color{magenta}3(x - 2) & = \color{magenta}6 \\ x - 2 & = \frac{\cancel6^2}{\cancel3_1} \\ x - 2 & = 2 \\ x & = 2 + 2 \\ \color{red} x & =\color{red} 4 \end{align*}

6. Which of the following statements are true?

(i) \( \displaystyle \left(0.6\right)^8 \div \left(0.6\right)^7 = \left(0.6\right)^2 \)

\begin{align*} & = \left(0.6\right)^8 \div \left(0.6\right)^7 \\ &= \left(0.6\right)^{8-7} \\ &= \left(0.6\right)^1 \\ \\ \left(0.6\right)^8 \div \left(0.6\right)^7 &\neq \left(0.6\right)^2 \end{align*}

Answer False

(ii) \( \displaystyle \left(\frac{12}{13}\right)^6 \div \left(\frac{12}{13}\right)^3 = \left(\frac{12}{13}\right)^3 \)

\begin{align*} & = \left(\frac{12}{13}\right)^6 \div \left(\frac{12}{13}\right)^3 \\ \\ &= \left(\frac{12}{13}\right)^{6-3} \\ \\ \left(\frac{12}{13}\right)^6 \div \left(\frac{12}{13}\right)^3 &= \left(\frac{12}{13}\right)^3 \\ \end{align*}

Answer True

(iii) The reciprocal of \( \displaystyle \left(\frac{7}{5}\right)^{12} \text{ is } \left(\frac{5}{7}\right)^{12} \)

\begin{align*} \text{Reciprocal of }\left(\frac{7}{5}\right)^{12} & = \left(\frac{5}{7}\right)^{12} \\ \end{align*}

Answer True

(iv) \( \displaystyle (5 + 5)^5 = 5^5 + 5^5 \)

\begin{align*} \text{LHS} & = (5 + 5)^5 \\ &= 10^5 \\ &= 100000 \\ \\ \text{RHS} &= 5^5 + 5^5 \\ &= 3125 + 3125 \\ &= 6250 \\ \\ (5 + 5)^5 &\neq 5^5 + 5^5 \end{align*}

Answer False

(v) \( \displaystyle \left[\left(\frac{1}{4}\right)^4 \div \left(\frac{1}{4}\right)^3\right] \div \left(\frac{1}{4}\right) = \frac{1}{4} \)

\begin{align*} \text{LHS} & = \left[\left(\frac{1}{4}\right)^4 \div \left(\frac{1}{4}\right)^3\right] \div \left(\frac{1}{4}\right) \\ \\ & = \left[\left(\frac{1}{4}\right)^{4-3}\right] \div \left(\frac{1}{4}\right) \\ \\ & = \left[\left(\frac{1}{4}\right)^1\right] \div \left(\frac{1}{4}\right) \\ \\ & = \left(\frac{1}{4}\right)^{1-1} \\ \\ & = \left(\frac{1}{4}\right)^0 \\ \\ & = 1 \\ \\ \text{RHS} &= \frac{1}{4} \\ \\ &\left[\left(\frac{1}{4}\right)^4 \div \left(\frac{1}{4}\right)^3\right] \div \left(\frac{1}{4}\right) \neq \frac{1}{4} \\ \\ \end{align*}

Answer False

(vi) \( \displaystyle \left(\frac{1}{7} \times \frac{1}{7^2}\right) \div \left(\frac{1}{7^3}\right) = 1 \)

\begin{align*} & = \left(\frac{1}{7} \times \frac{1}{7^2}\right) \div \frac{1}{7^3} \\ \\ & = \left(\frac{1}{7^{1+2}} \right) \div \frac{1}{7^3} \\ \\ & = \frac{1}{7^3} \div \frac{1}{7^3} \\ \\ & = \frac{1}{7^{3-3}} \\ \\ & = \frac{1}{7^{0}} \\ \\ & = \frac{1}{1} \\ \\ & = 1 \end{align*}

Answer True