\begin{array}{l}
\text {Q1. Using laws of exponents, simplify and write the answer in exponential form: } \\
\text {(i) } \quad 2^{3} \times 2^{4} \times 2^{5} \quad
\text {(ii) } \quad 5^{12}+5^{3} \quad
\text {(iii) } \quad (7^{2})^{3} \\
\text {(iv) } \quad (3^{2})^{5} \div 3^{4} \quad
\text {(v) } \quad 3^{7} \times 2^{7} \quad
\text {(vi) } \quad (5^{21} \div 5^{13}) \times 5^{7}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {(i) We know that } a^{m} \times a^{n} \times a^{p} =a^{(m+n+p)} \\
\Rightarrow \quad 2^{3} \times 2^{4} \times 2^{5}=2^{(3+4+5)}=2^{12} \\ \\\text {(ii) We know that } a^{m} \div a^{n}=a^{m-n} \\
\Rightarrow \quad 5^{12} \div 5^{3}=5^{12-3}=5^{9} \\ \\\text {(iii) We know that } (a^{m})^{n}=a^{mn} \\
\Rightarrow \quad (7^{2})^{3}=7^{2 \times 3}= 7^{6} \\ \\\text {(iv) } \quad (3^{2})^{5} \div 3^{4} \\
=3^{2 \times 5} \div 3^{4} \quad \quad [\because (a^{m})^{n}=a^{mn}] \\
=3^{10} \div 3^{4} \\
=3^{(10-4)} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
=3^{6} \\ \\\text {(v) } \quad 3^{7} \times 2^{7} \\
=(3 \times 2)^{7} \quad \quad [\because a^{m} \times b^{m}=(a \times b)^{m}] \\
=6^{7} \\ \\\text {(vi) } \quad (5^{21} \div 5^{13}) \times 5^{7} \\
=5^{(21 – 13)} \times 5^{7} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
=5^{8} \times 5^{7} \quad \quad [\because a^{m} \times a^{n} =a^{(m+n)}] \\
=5^{8+7} \\
=5^{15} \\\end{array}\begin{array}{l}
\text {Q2. Simplify and express each of the following in exponential form: } \\
\text {(i) } \quad \{(2^{3})^{4} \times 2^{8}\} \div 2^{12} \quad
\text {(ii) } \quad (8^{2} \times 8^{4}) \div 8^{3} \\
\text {(iii) } \quad (\frac{5^{7}}{5^{2}}) \times 5^{3} \quad
\text {(iv) } \quad \frac{5^{4} \times x^{10} y^{5}}{5^{4} \times x^{7} y^{4}}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {(i) } \quad \{(2^{3})^{4} \times 2^{8}\} \div 2^{12} \\\{(2^{3})^{4} \times 2^{8}\} \div 2^{12} \\
=\{2^{12} \times 2^{8}\} \div 2^{12} \quad \quad [\because (a^{m})^{n}=a^{m n}] \\
=2^{(12+8)} \div 2^{12} \quad \quad [\because a^{m} \times a^{n}=a^{(m+n)}] \\
=2^{20} \div 2^{12} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
=2^{(20-12)} \\
=2^{8} \\ \\\text {(ii) } \quad (8^{2} \times 8^{4}) \div 8^{3} \\
=(8^{2} \times 8^{4}) \div 8^{3} \quad \quad [\because a^{m} \times a^{n}=a^{(m+n)}] \\
=8^{(2+4)} \div 8^{3} \\
=8^{6} \div 8^{3} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
=8^{(6-3)} = 8^{3} \\
=(2^{3})^{3}=2^{9} \\ \\\text {(iii) } \quad (\frac{5^{7}}{5^{2}}) \times 5^{3} \\
=5^{(7-2)} \times 5^{3} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\=5^{5} \times 5^{3} \quad \quad [\because a^{m} \times a^{n}=a^{(m+n)}] \\
=5^{(5+3)}=5^{8} \\ \\\text {(iv) } \quad \frac{5^{4} \times x^{10} y^{5}}{5^{4} \times x^{7} y^{4}} \\
=(5^{4-4} \times x^{10-7} y^{5-4}) \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
=5^{0} x^{3} y^{1} \quad \quad [\because 5^{0}=1] \\
=x^{3} y\end{array}\begin{array}{l}
\text {Q3. Simplify and express each of the following in exponential form: } \\
\text {(i) } \quad \{(3^{2})^{3} \times 2^{6}\} \times 5^{6} \quad
\text {(ii) } \quad (\frac{x}{y})^{12} \times y^{24} \times (2^{3})^{4} \\
\text {(iii) } \quad (\frac{5}{2})^{6} \times (\frac{5}{2})^{2} \quad
\text {(iv) } \quad (\frac{2}{3})^{5} \times (\frac{3}{5})^{5}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {(i) } \quad \{(3^{2})^{3} \times 2^{6}\} \times 5^{6} \\=\{3^{6} \times 2^{6}\} \times 5^{6} \quad \quad [\because (a^{m})^{n}=a^{m n}] \\
=6^{6} \times 5^{6} \quad \quad [\because a^{m} \times b^{m}=(a \times b)^{m}] \\
=30^{6} \\ \\\text {(ii) } \quad (\frac{x}{y})^{12} \times y^{24} \times (2^{3})^{4} \\
=(\frac {x^{12}}{y^{12}}) \times y^{24} \times 2^{12} \\
=x^{12} \times y^{24-12} \times 2^{12} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
=x^{12} \times y^{12} \times 2^{12} \\
=(2xy)^{12} \quad \quad [\because a^{m} \times b^{m} \times c^{m}=(a \times b \times c)^{m}] \\ \\\text {(iii) } \quad (\frac{5}{2})^{6} \times (\frac{5}{2})^{2} \\
=(\frac{5}{2})^{6+2} \quad \quad [\because a^{m} \times a^{n}=a^{(m+n)}] \\
=(\frac{5}{2})^{8} \\ \\\text {(iv) } \quad (\frac{2}{3})^{5} \times (\frac{3}{5})^{5} \\
=(\frac {2 \times 3}{3 \times 5})^{5} \quad \quad [\because a^{m} \times b^{m}=(a \times b)^{m}] \\
=\frac {2}{5}^{5}
\end{array}\begin{array}{l}
\text {Q4. Write } 9 \times 9 \times 9 \times 9 \times 9 \text { in exponential form with base 3.}
\end{array}\begin{array}{l}
\text {Sol. } \\9 \times 9 \times 9 \times 9 \times 9=9^{5} \\
=(3^{2})^{5} \quad \quad [\because 9 = 3^{2}] \\
=3^{10}\end{array}\begin{array}{l}
\text {Q5. Simplify and write each of the following in exponential form: } \\
\text {(i) } \quad (25)^{3} \div 5^{3} \quad
\text {(ii) } \quad (81)^{5} \div (3^{2})^{5} \\
\text {(iii) } \quad \frac{9^{8} \times (x^{2})^{5}}{(27)^{4} \times (x^{3})^{2}} \quad
\text {(iv } \quad \frac{3^{2} \times 7^{8} \times 13^{6} }{21^{2} \times 91^{3}}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {(i) } \quad (25)^{3} \div 5^{3} \\
=(5^{2})^{3} \div 5^{3} \\
=5^{2 \times 3} \div 5^{3} \quad \quad [\because (a^{m})^{n}=a^{m n}] \\
=5^{6} \div 5^{3} \\
=5^{6-3} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
=5^{3} \\ \\\text {(ii) } \quad (81)^{5} \div (3^{2})^{5} \\
=(81)^{5} \div 3^{10} \quad \quad [\because (a^{m})^{n}=a^{m n}] \\
=(3^{4})^{5} \div 3^{10} \quad \quad [\because 81=3^{4}] \\
=3^{20} \div 3^{10} \quad \quad [\because (a^{m})^{n}=a^{m n}] \\
=3^{20-10} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
=3^{10} \\ \\\text {(iii) } \quad \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}} \quad \quad [\because (a^{m})^{n}=a^{m n}] \\
= 3^{16-12} \times x^{10 – 6} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
=3^{4} \times x^{4} \\
=(3x)^{4} \\ \\\text {(iv } \quad \frac{3^{2} \times 7^{8} \times 13^{6} }{21^{2} \times 91^{3}} \\
= \frac{3^{2} \times 7^{8} \times 13^{6} }{(7 \times 3)^{2} \times (13 \times 7)^{3}} \\
= \frac{3^{2} \times 7^{8} \times 13^{6} }{7^{2} \times 3^{2} \times 13^{3} \times 7^{3}} \quad \quad [\because (ab)^{m}=a^{m} \times b^{n}] \\
= 7^{(8-2-3)} \times 13^{(6-3)} \quad \quad [\because a^{m} \div a^{n}=a^{m-n}] \\
= 7^{3} \times 13^{3} \\
= (7 \times 13)^{3} \\
= 91^{3} \\
\end{array}\begin{array}{l}
\text {Q6. Simplify: } \\\text {(i) } \quad (3^{5})^{11} \times (3^{15})^{4}-(3^{5})^{18} \times(3^{5})^{5} \quad
\text {(ii) } \quad \frac{16 \times 2^{n+1}-4 \times 2^{n}}{16 \times 2^{n+2}-2 \times 2^{n+2}} \\
\text {(iii) } \quad \frac{10 \times 5^{n+1}+25 \times 5^{n}}{3 \times 5^{n+2}+10 \times 5^{n+1}} \quad
\text {(iv) } \quad \frac{(16)^{7} \times(25)^{5} \times(81)^{3}}{(15)^{7} \times(24)^{5} \times(80)^{3}}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {(i) } \quad (3^{5})^{11} \times (3^{15})^{4}-(3^{5})^{18} \times(3^{5})^{5} \\
=(3)^{5 \times 11} \times(3)^{15 \times 4}-(3)^{5 \times 18} \times(3)^{5 \times 5} \quad \quad [\because (a^{m})^{n}=a^{m n}] \\
=3^{55} \times 3^{60} – 3^{90} \times 3^{25} \\
=3^{55+60} – 3^{90+25} \\
=3^{115}-3^{115} = 0 \\ \\\text {(ii) } \quad \frac{16 \times 2^{n+1}-4 \times 2^{n}}{16 \times 2^{n+2}-2 \times 2^{n+2}} \\
=\frac{16 \times 2^{n} \times 2 – 4 \times 2^{n}}{16 \times 2^{n+2} – 2 \times 2^{n+2}} \\
= \frac{4 \times 2^{n} \times (4 \times 2 – 1)}{2 \times 2^{n+2} \times (8 -1)} \\
= \frac{2^{2} \times 2^{n} \times 7}{2 \times 2^{n+2} \times 7} \\
= \frac{2^{n+2} \times 7}{2 \times 2^{n+2} \times 7} \\
= \frac{1}{2} \\ \\\text {(iii) } \quad \frac{10 \times 5^{n+1}+25 \times 5^{n}}{3 \times 5^{n+2}+10 \times 5^{n+1}} \\
=\frac{2 \times 5 \times 5^{n+1} + 5^{2} \times 5^{n}}{3 \times 5^{n+2} + 2 \times 5 \times 5^{n+1}} \\
=\frac{2 \times 5^{n+1+1} + 5^{n +2}}{3 \times 5^{n+2} + 2 \times 5^{n+1+1}} \\
=\frac{5^{n+2} \times (2+1)}{5^{n+2} \times (3+2)} \\
=\frac{3}{5} \\ \\\text {(iv) } \quad \frac{(16)^{7} \times(25)^{5} \times(81)^{3}}{(15)^{7} \times(24)^{5} \times(80)^{3}} \\ \\=\frac{(2 \times 2 \times 2 \times 2)^{7} \times (5 \times 5)^{5} \times(33 \times 3 \times 3 \times 3)^{3}}{(3 \times 5)^{7} \times(2 \times 2 \times 2 \times 3)^{5} \times(2 \times 2 \times 2 \times 2 \times 5)^{3}} \\ \\=\frac{(2^{4})^{7} \times (5^{2})^{5} \times(3^{4})^{3}}{3^{7} \times 5^{7} \times (2^{3} \times 3)^{5} \times (2^{4} \times 5)^{3}} \\ \\=\frac{2^{28} \times 5^{10} \times 3^{12}}{3^{7} \times 5^{7} \times 2^{15} \times 3^{5} \times 2^{12} \times 5^{3}} \\
=2^{28-15-12} \times 5^{10-3-7} \times 3^{12-7-5} \\
=2^{1} \times 5^{0} \times 3^{0} \\
=2 \quad \quad [\because a^{0} = 1] \\ \\\end{array}\begin{array}{l}
\text {Q7. Find the values of n in each of the following:} \\
\text {(i) } \quad 5^{2 n} \times 5^{3}=5^{11} \quad
\text {(ii) } \quad 9 \times 3^{n}=3^{7} \\
\text {(iii) } \quad 8 \times 2^{n+2}=32 \quad
\text {(iv) } \quad 7^{2 n+1} \div 49=7^{3} \quad
\text {(v) } \quad (\frac{3}{2})^{4} \times (\frac{3}{2})^{5}=(\frac{3}{2})^{2 n+1} \quad
\text {(vi) } \quad (\frac{2}{3})^{10} \times \{(\frac{3}{2})^{2}\}^{5}=(\frac{2}{3})^{2 n-2}\end{array}\begin{array}{l}
\text {Sol. } \\\text {(i) } \quad 5^{2 n} \times 5^{3}=5^{11} \\
\Rightarrow \quad 5^{(2n+3)}=5^{11} \\
\text {On equating the coefficients, we get } \\
2n+3=11 \\
\Rightarrow \quad 2 n=11-3 \\
\Rightarrow \quad 2 n=8 \\
\Rightarrow \quad n=\frac {8}{2} \\
\Rightarrow \quad n=4 \\ \\\text {(ii) } \quad 9 \times 3^{n}=3^{7} \\
\Rightarrow \quad (3)^{2} \times 3^{n}=3^{7} \\
\Rightarrow \quad 3^{(2+n)}=3^{7} \\
\text {On equating the coefficients, we get } \\
2+n=7 \\
\Rightarrow \quad n=7-2=5 \\ \\\text {(iii) } \quad 8 \times 2^{(n+2)}=32 \\
\Rightarrow \quad (2)^{3} \times 2^{(n+2)}=(2)^{5} \quad \quad [\because 8 =2^{3} \text { and } 32 =2^{5}] \\
\Rightarrow \quad 2^{(3+n+2)}=(2)^{5} \\
\text {On equating the coefficients, we get } \\
3+n+2=5 \\
\Rightarrow \quad n+5=5 \\
\Rightarrow \quad n=0 \\ \\\text {(iv) } \quad 7^{(2n+1)} \div 49=7^{3} \\
\Rightarrow \quad 7^{(2n+1)} \div 7^{2}=7^{3} \quad \quad [\because 49 =7^{2}] \\
\Rightarrow \quad 7^{(2n+1-2)}=7^{3} \\
\Rightarrow \quad 7^{2n-1}=7^{3} \\
\text {On equating the coefficients, we get } \\
2n-1=3 \\
\Rightarrow \quad n=4 \\
\Rightarrow \quad n=2 \\ \\\text {(v) } \quad (\frac{3}{2})^{4} \times (\frac{3}{2})^{5}=(\frac{3}{2})^{2n+1} \\
\Rightarrow \quad (\frac{3}{2})^{4+5} =(\frac{3}{2})^{2n+1} \\
\Rightarrow \quad (\frac{3}{2})^{9} =(\frac{3}{2})^{2n+1} \\
\text {On equating the coefficients, we get } \\
2n+1=9 \\
\Rightarrow \quad 2n=9-1 \\
\Rightarrow \quad n =4 \\ \\\text {(vi) } \quad (\frac{2}{3})^{10} \times \{(\frac{3}{2})^{2}\}^{5}=(\frac{2}{3})^{2 n-2} \\
\Rightarrow \quad (\frac{2}{3})^{10} \times (\frac{3}{2})^{10}=(\frac{2}{3})^{2 n-2} \\
\Rightarrow \quad (\frac{2 \times 3}{3 \times 2})^{10} =(\frac{2}{3})^{2 n-2} \\
\Rightarrow \quad 1 =(\frac{2}{3})^{2 n-2} \\
\Rightarrow \quad (\frac{2}{3})^{0} =(\frac{2}{3})^{2n-2} \\\text {On equating the coefficients, we get } \\
2 n-2 = 0 \\
\Rightarrow \quad 2n=2 \\
\Rightarrow \quad n=1 \\\end{array}\begin{array}{l}
\text {Q8. If } \frac{9^{n} \times 3^{2} \times 3^{n}-(27)^{n}}{(3^{3})^{5} \times 2^{3}}=\frac{1}{27} \text {, find the value of n }
\end{array}\begin{array}{l}
\text {Sol. } \\\frac{9^{n} \times 3^{2} \times 3^{n}-(27)^{n}}{(3^{3})^{5} \times 2^{3}}=\frac{1}{27} \\
\Rightarrow \quad \frac{(9 \times 3)^{n} \times 3^{2} -(27)^{n}}{(3^{3})^{5} \times 2^{3}}=\frac{1}{27} \\
\Rightarrow \quad \frac{(27)^{n} \times 3^{2} -(27)^{n}}{(27)^{5} \times 2^{3}}=\frac{1}{27} \\
\Rightarrow \quad \frac{(27)^{n} \times (3^{2} -1)}{(27)^{5} \times 2^{3}}=\frac{1}{27} \\
\Rightarrow \quad \frac{(27)^{n} \times 8}{(27)^{5} \times 2^{3}}=\frac{1}{27} \\
\Rightarrow \quad \frac{(27)^{n}}{(27)^{5}}=\frac{1}{27} \\
\Rightarrow \quad 27^{(n-5)}=27^{-1} \\\text {On equating the coefficients, we get } \\
n-5=-1 \\
\Rightarrow \quad n = -1+5 = 4 \\\end{array}
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