Mathematics Solutions Solutions for Class 7 Math Chapter 6 - Indices

Answer:

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) 2^{10}=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\ =1024 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) 5^3=5\times 5\times 5 \\ =125 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) {\left(-7\right)}^4 =\left(-7\right)\times \left(-7\right)\times \left(-7\right)\times \left(-7\right) \\ =2401 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right) {\left(-6\right)}^3=\left(-6\right)\times \left(-6\right)\times \left(-6\right) \\ =-216 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right) 9^3=9\times 9\times 9 \\ = 729 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vi}\right) 8^1=8 \\ =8 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vii}\right) {\left(\frac{4}{5}\right)}^3=\frac{4}{5}\times \frac{4}{5}\times \frac{4}{5} \\ =\frac{64}{125} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{viii}\right) {\left(-\frac{1}{2}\right)}^4=\left(-\frac{1}{2}\right)\times \left(-\frac{1}{2}\right)\times \left(-\frac{1}{2}\right)\times \left(-\frac{1}{2}\right) \\ =\frac{1}{16} \end{gathered}$$

Answer:

It is known that, a^m × aⁿ = a^m+n, where m and n are integers and a is a non-zero rational number.
$$\begin{gathered} \left(\mathrm{i}\right) 7^4\times 7^2 \\ = 7^{4+2} \\ =7^6 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) {\left(-11\right)}^5\times {\left(-11\right)}^2 \\ ={\left(-11\right)}^{5+2} \\ ={\left(-11\right)}^7 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) {\left(\frac{6}{7}\right)}^3\times {\left(\frac{6}{7}\right)}^5 \\ ={\left(\frac{6}{7}\right)}^{3+5} \\ ={\left(\frac{6}{7}\right)}^8 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right) {\left(-\frac{3}{2}\right)}^5\times {\left(-\frac{3}{2}\right)}^3 \\ ={\left(-\frac{3}{2}\right)}^{5+3} \\ ={\left(-\frac{3}{2}\right)}^8 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right) a^{16}\times a^7 \\ =a^{16+7} \\ =a^{23} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vi}\right) {\left(\frac{p}{5}\right)}^3\times {\left(\frac{p}{5}\right)}^7 \\ ={\left(\frac{p}{5}\right)}^{3+7} \\ ={\left(\frac{p}{5}\right)}^{10} \end{gathered}$$

Answer:

It is known that, a^m ÷ aⁿ = a^m−n, where m and n are integers and a is non-zero rational number.
$$\begin{gathered} \left(\mathrm{i}\right) a^6÷a^4 \\ =a^{6-4} \\ =a^2 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) m^5÷m^8 \\ =m^{5-8} \\ =m^{-3} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) p^3÷p^{13} \\ =p^{3-13} \\ =p^{-10} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right) x^{10}÷x^{10} \\ =x^{10-10} \\ =x^0 \\ =1 \left(\because a^0=1\right) \end{gathered}$$

Answer:

It is known that, a^m ÷ aⁿ = a^m−n, where m and n are integers and a is a non-zero rational number.
$$\begin{gathered} \left(\mathrm{i}\right) {\left(-7\right)}^{12}÷{\left(-7\right)}^{12} \\ ={\left(-7\right)}^{12-12} \\ ={\left(-7\right)}^0 \\ =1 \left(\because a^0=1\right) \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) 7^5÷7^3 \\ =7^{5-3} \\ =7^2 \\ =7\times 7 \\ =49 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) {\left(\frac{4}{5}\right)}^3÷{\left(\frac{4}{5}\right)}^2 \\ ={\left(\frac{4}{5}\right)}^{3-2} \\ ={\left(\frac{4}{5}\right)}^1 \\ =\frac{4}{5} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right) 4^7÷4^5 \\ =4^{7-5} \\ =4^2 \\ =4\times 4 \\ =16 \end{gathered}$$

Answer:

It is known that, (a^m)ⁿ = a^mn, where m and n are integers and a is a non-zero rational number.
$$\begin{gathered} \left(\mathrm{i}\right) {\left[{\left(\frac{15}{12}\right)}^3\right]}^4 \\ ={\left(\frac{15}{12}\right)}^{3\times 4} \\ ={\left(\frac{15}{12}\right)}^{12} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) {\left(3^4\right)}^{-2} \\ =3^{4\times \left(-2\right)} \\ =3^{-8} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) {\left[{\left(\frac{1}{7}\right)}^{-3}\right]}^4 \\ ={\left(\frac{1}{7}\right)}^{-3\times 4} \\ ={\left(\frac{1}{7}\right)}^{-12} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right) {\left[{\left(\frac{2}{5}\right)}^{-2}\right]}^{-3} \\ ={\left(\frac{2}{5}\right)}^{\left(-2\right)\times \left(-3\right)} \\ ={\left(\frac{2}{5}\right)}^6 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right) {\left(6^5\right)}^4 \\ =6^{5\times 4} \\ =6^{20} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vi}\right) {\left[{\left(\frac{6}{7}\right)}^5\right]}^2 \\ ={\left(\frac{6}{7}\right)}^{5\times 2} \\ ={\left(\frac{6}{7}\right)}^{10} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vii}\right) {\left[{\left(\frac{2}{3}\right)}^{-4}\right]}^5 \\ ={\left(\frac{2}{3}\right)}^{\left(-4\right)\times 5} \\ ={\left(\frac{2}{3}\right)}^{-20} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{viii}\right) {\left[{\left(\frac{5}{8}\right)}^3\right]}^{-2} \\ ={\left(\frac{5}{8}\right)}^{3\times \left(-2\right)} \\ ={\left(\frac{5}{8}\right)}^{-6} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ix}\right) {\left[{\left(\frac{3}{4}\right)}^6\right]}^1 \\ ={\left(\frac{3}{4}\right)}^{6\times 1} \\ ={\left(\frac{3}{4}\right)}^6 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{x}\right) {\left[{\left(\frac{2}{5}\right)}^{-3}\right]}^2 \\ ={\left(\frac{2}{5}\right)}^{\left(-3\right)\times 2} \\ ={\left(\frac{2}{5}\right)}^{-6} \end{gathered}$$

Answer:

It is known that, $a^{-m}=\frac{1}{a^m}$ where m is an integer and a is a non-zero rational number.
$$\begin{gathered} \left(\mathrm{i}\right) {\left(\frac{2}{7}\right)}^{-2} \\ =\frac{1}{{\left({\displaystyle \frac{2}{7}}\right)}^2} \\ ={\left(\frac{7}{2}\right)}^2 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) {\left(\frac{11}{3}\right)}^{-5} \\ =\frac{1}{{\left({\displaystyle \frac{11}{3}}\right)}^5} \\ ={\left(\frac{3}{11}\right)}^5 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) {\left(\frac{1}{6}\right)}^{-3} \\ =\frac{1}{{\left({\displaystyle \frac{1}{6}}\right)}^3} \\ ={\left(\frac{6}{1}\right)}^3 \\ =6^3 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right) {\left(y\right)}^{-4} \\ =\frac{1}{y^4} \\ ={\left(\frac{1}{y}\right)}^4 \end{gathered}$$

Answer:

(i) The prime factorization of 625 is,
625 = 5 × 5 ×  5 × 5
To find the square root, we will take one number from each pair and multiply.
$$\begin{gathered} \sqrt{625}=5\times 5=25 \\ \therefore \sqrt{625}=25 \end{gathered}$$
(ii) The prime factorization of 1225 is,
1225 = 5 × 5 × 7 × 7
To find the square root, we will take one number from each pair and multiply.
$$\begin{gathered} \sqrt{1225}=5\times 7=35 \\ \therefore \sqrt{1225}=35 \end{gathered}$$
(iii) The prime factorisation of 289 is,
289 = 17 × 17
To find the square root, we will take one number from each pair and multiply.
$$\begin{gathered} \sqrt{289}=17\times 17=17 \\ \therefore \sqrt{289}=17 \end{gathered}$$
(iv) The prime factorization of 4096 is,
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 â€‹× 2
To find the square root, we will take one number from each pair and multiply.
$$\begin{gathered} \sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2 = 64 \\ \therefore \sqrt{4096}=64 \end{gathered}$$
(v) The prime factorizationn of 1089 is,
1089 = 3 × 3 × 11 × 11
To find the square root, we will take one number from each pair and multipy.
$$\begin{gathered} \sqrt{1089}=3\times 11=33 \\ \therefore \sqrt{1089}=33 \end{gathered}$$