Exercise 2.1
Q1 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Question 1
(i) $\left(\frac{3}{5}\right)^{-2}$
(ii) $(-3)^{-3}$
(iii) $\left(\frac{2}{7}\right)^{-4}$
Sol :
(i)
$\begin{aligned}\left(\frac{3}{5}\right)^{2} &=\left(\frac{5}{3}\right)^{2} \\ &=\frac{5 \times 5}{3 \times 3} \\ &=\frac{25}{9} \end{aligned}$ $\left(\because(p / q)^{-m}=\left(\frac{q}{p}\right)^{m}\right)$
(ii)
$\begin{aligned}(-3)^{-3} &=\frac{1}{(-3)^{3}} \quad\left(\because(a)^{-m}=\frac{1}{(a)^{m}}\right) \\ &=\frac{1}{(-3) \times(-3) \times(-3)} \\ &=\frac{1}{-27}=\frac{-1}{27} \end{aligned}$
(iii)
$\begin{aligned}\left(\frac{2}{7}\right)^{-4} &=\left(\frac{7}{2}\right)^{4} \\ &=\frac{7 \times 7 \times 7 \times 7}{2 \times 2 \times 2 \times 2} \\ &=\frac{2401}{16} \end{aligned}$
Q2 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Simplify:
(i) $[(2)^{-1}+(4)^{-1}+(3)^{-1}]^{-1}$
(ii) $[(4)^{-1}-(5)^{-1}]^2 \times \left( \frac{5}{8}\right)^{-1}$
(iii) $[4^{0}+4^{2}-2^3] \times 3^{-2}$
(iv) $\left[(5)^2-\left(\frac{1}{4}\right)^{-2}\right] \times \left(\frac{3}{4}\right)^{-2}$
Sol :
(i)
$\begin{aligned}\left[(2)^{-1}+(4)^{-1}+(3)^{-1}\right]^{-1} &=\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{3}\right]^{-1} \quad\left(\because(a)^{-m}=\frac{1}{a^{m}}\right) \\ &=\left[\frac{6+3+4}{12}\right]^{-1} \\ &=\left(\frac{13}{12}\right)^{-1} \\ &=\frac{12}{13} \end{aligned}$
(ii)
$\begin{aligned}\left[(4)^{-1}-(5)^{-1}\right]^{2} \times\left(\frac{5}{8}\right)^{-1} &=\left(\frac{1}{4}-\frac{1}{5}\right)^{2} \times\left(\frac{8}{5}\right)^{1} \\ &=\left(\frac{5-4}{20}\right)^{2} \times \frac{8}{5} \\ &=\frac{1}{20 \times 20} \times \frac{8}{5} \\ &=\frac{1}{250} \end{aligned}$
(iii)
$\begin{aligned}\left[4^{0}+4^{2}-2^{3}\right] \times 3^{-2} &=[1+16-8] \times \frac{1}{3^{2}} \\ &=(9) \times \frac{1}{9} . \\ &=1 . \end{aligned}$
(iv)
$\begin{aligned}\left[(5)^{2}-\left(\frac{1}{4}\right)^{-2}\right] \times\left(\frac{3}{4}\right)^{-2} &=\left[(25)-(4)^{2}\right] \times\left(\frac{4}{3}\right)^{2} \\ &=[25-16] \times \frac{16}{9} \\ &=9 \times \frac{16}{9} \\ &=16 . \end{aligned}$
Q3 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Find the multiplicative inverse of the following:
(i) $\left(\dfrac{81}{16}\right)^{\frac{-3}{4}}$
(ii) $\left\{ \left(\dfrac{-3}{2}\right)^{-4} \right\}^{\frac{1}{2}}$
(iii) $\left(\frac{5}{7}\right)^{-2} \times \left( \frac{5}{7}\right)^{4} \div \left(\frac{5}{7}\right)^{3}$
Sol :
(i)
$=\left(\dfrac{81}{16}\right)^{\frac{-3}{4}}=\left(\dfrac{16}{81}\right)^{\frac{3}{4}}$
$=\left(\dfrac{2^4}{3^4}\right)^{\frac{3}{4}}=\left(\dfrac{2}{3}\right)^{4 \times \frac{3}{4}}$
$=\left(\dfrac{2}{3}\right)^{3}=\dfrac{2}{3}\times \dfrac{2}{3}\times\dfrac{2}{3}$
$=\dfrac{8}{27}$
∴Multiplicative Inverse of $=\dfrac{8}{27}=\dfrac{27}{8}$
(ii)
$=\left\{\left(\dfrac{-3}{2}\right)^{-4}\right\}^{\frac{1}{2}}=\left(\dfrac{-3}{2}\right)^{-4 \times \frac{1}{2}}$
$=\left(\dfrac{-3}{2}\right)^{-2}=\left(\frac{2}{-3}\right)^{2}=\dfrac{2 \times 2}{-3 \times -3}=\dfrac{4}{9}$
∴Multiplicative Inverse of $\dfrac{4}{9}=\dfrac{9}{4}$
(iii)
$=\left(\dfrac{5}{7}\right)^{-2} \times \left(\dfrac{5}{7}\right)^{4} \div \left(\dfrac{5}{7}\right)^{3}$
$=\left( \dfrac{7}{5} \right)^{2} \times \left(\dfrac{5}{7} \right)^{4} \div \left(\dfrac{5}{7}\right)^3$
$=\left(\dfrac{7}{5}\right)^{2} \times \left(\dfrac{5}{7}\right)^{4-3}=\left(\dfrac{7}{5}\right)^{2} \times \left(\dfrac{5}{7}\right)^{1}$
$=\dfrac{7}{5} \times\dfrac{7}{5} \times\dfrac{7}{5} =\dfrac{7}{5} $
∴Multiplicative Inverse of $\dfrac{7}{5}=\frac{5}{7}$
Q4 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
(i) Express $16^{-2}$ as a power with base 2.
(ii) Express $125^{-4}$ as a power with base 5.
Sol :
(i)
$\begin{array}{rlr}(16)^{-2} & =(2 \times 2 \times 2 \times 2)^{-2} & \\ & =\left(2^{4}\right)^{-2} & \\ & =2^{4 \times -2} & \left(\because\left(a^{m}\right)^{n}=a^{m n}\right) \\ & =2^{-8}\\ & =\left(\frac{1}{2}\right)^{8}=\frac{1}{2^{8}} .\end{array}$
(ii)
$\begin{aligned}(125)^{-4} &=(5 \times 5 \times 5)^{-4} \\ &=\left((5)^{3}\right)^{-4} \\ &=5^{3 \times(-4)} \\ &=5^{(-12)} \\ &=\left(\dfrac{1}{5}\right)^{12}=\frac{1}{5^{12}} \end{aligned}$
Q5 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Write the following numbers in expanded form using exponents:
(i) 2789.453
(ii) 3007.805
Sol :
(i)
$\begin{aligned} 2789.453 &=2 \times 1000+7 \times 100+8 \times 10+9 \times 1+\frac{4}{10}+\frac{5}{100}+\frac{3}{1000} \\ &=2 \times 10^{3}+7 \times 10^{2}+8 \times 10^{1}+9 \times 10^{\circ}+4 \times 10^{-1}+5 \times 10^{-2}+3 \times 10^{-3} \end{aligned}$
(ii) 3007.805
$=3 \times 1000+0 \times 100+0 \times 10+7 \times 1+\frac{8}{10}+\frac{0}{100}+\frac{5}{1000}$
$=3 \times 10^{3}+0 \times 10^{2}+0 \times 10^{1}+7 \times 10^{\circ}+8 \times 10^{-1}+0 \times 10^{-2}+5 \times 10^{-3}$
Q6 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Simplify and write in exponential form with positive exponent:
(i) $\left[\left(\left(\frac{5}{7}\right)^{2}\right)^{-1}\right]^{-3}$
(ii) $\left(\frac{2}{7}\right)^{2} \times\left(\frac{7}{2}\right)^{-3} \div\left\{\left(\frac{7}{5}\right)^{-2}\right\}^{-4}$
(iii) $\left(\frac{4}{5}\right)^{2} \times(5)^{4} \times \left(\frac{2}{5}\right)^{-2} \div \left(\frac{5}{2}\right)^{-3}$
(iv) $ \frac{8^{-1} \times 5^{3}}{2^{-4}}$
Sol :
(i)
$\begin{aligned}\left[\left(\left(\frac{5}{7}\right)^{2}\right)^{-1}\right]^{-3} &=\left(\frac{5}{7}\right)^{2 x-1 x-3} \quad\left(\because\left(a^{m}\right)^{n}=a^{m \times n}\right) \\ &=\left(\frac{5}{7}\right)^{6} \end{aligned}$
(ii)
$\left(\frac{2}{7}\right)^{2} \times\left(\frac{7}{2}\right)^{-3} \div\left\{\left(\frac{7}{5}\right)^{-2}\right\}^{-4}$
$=\left(\frac{2}{7}\right)^{2} \times\left(\frac{2}{7}\right)^{3} \div\left(\frac{7}{5}\right)^{(-2) \times(-4)} \quad\left[\because\left(\frac{p}{q}\right)^{-m}=\left(\frac{q}{p}\right)^{m}\right]$
$=\left(\frac{2}{7}\right)^{5} \div\left(\frac{7}{5}\right)^{8}$
$=\left(\frac{2}{7}\right)^{5} \times\left(\frac{5}{7}\right)^{8}$
$=\frac{2^{5} \times 5^{8}}{7^{5} \times 7^{8}}=\frac{2^{5} \times 5^{8}}{7^{(5+8)}} \quad\left(\because a^{m} \times a^{n}=a^{m+n}\right)$
$=\frac{2^{5} \times 5^{8}}{7^{13}}$
(iii)
$\begin{aligned}\left(\frac{4}{5}\right)^{2} & \times(5)^{4} \times\left(\frac{2}{5}\right)^{-2} \div\left(\frac{5}{2}\right)^{-3} \\ &=\left(\frac{4}{5}\right)^{2} \times(5)^{4} \times\left(\frac{5}{2}\right)^{2} \div\left(\frac{2}{5}\right)^{3} \\ &=\frac{4^{2} \times 5^{4}}{5^{2}} \times \frac{5^{2}}{2^{2}} \times \frac{5^{3}}{2^{3}} \\ &=\frac{5^{4+2+3-2} \times\left(2^{4}\right)^{2}}{2^{2+3}}=\frac{5^{7}}{2^{5-4}}=\frac{5^{7}}{2^{1}} . \end{aligned}$
(iv)
$\begin{aligned} \frac{8^{-1} \times 5^{3}}{2^{-4}} &=\frac{\left(2^{3}\right)^{-1} \times 5^{3}}{2^{-4}} \\ &=2^{-3+4} \times 5^{3} \\ &=5^{3} \times 2^{1}\\ &=5\times 5 \times 5 \times 2=250 \end{aligned}$
Q7 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Simplify and write the following in exponential form:
(i) $\left((-2)^{3}\right)^{2}+(5)^{-3} \div(5)^{-5}-(-1 / 2)^{0}$
(ii) $3^{-5} \times 3^{2} \div 3^{-6}+\left(2^{2} \times 3^{2}\right)^{2}+\left(\frac{2}{3}\right)^{-1}+2^{-1}+\left(\frac{1}{19}\right)^{-1}$
Sol :
(i) $\left((-2)^{3}\right)^{2}+(5)^{-3} \div(5)^{-5}-(-1 / 2)^{0}$
$=(-2)^{3 \times 2}+(5)^{-3} \times(5)^{+5}-1 \quad\left(a^{0}=1 ;\left(a^{m}\right)^{n}=a^{m n}\right)$
$=2^{6}+5^{2}-1$
=64+25-1
=88
$=8 \times 11$
$=2^{3} \times 11$
(ii) $3^{-5} \times 3^{2} \div 3^{-6}+\left(2^{2} \times 3\right)^{2}+\left(\frac{2}{3}\right)^{-1}+2^{-1}+\left(\frac{1}{19}\right)^{-1}$
$=3^{-5+2} \times 3^{6}+2^{4} \times 3^{2}+\left(\frac{3}{2}\right)^{1}+\left(\frac{1}{2}\right)^{1}+(19)^{1}$
$=3^{-3+6}+2^{4} \times 3^{2}+\frac{3+1}{2}+19$
$=3^{3}+2^{4} \times 3^{2}+\frac{4}{2}+19$
$=3^{3}+16\times 9 +2+19$
=27+144+2+19
$=192=3 \times 64$
$=2^{6} \times 3$
Q8 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Simplify and write in exponential form with negative exponent:
(i) $5^{3} \times\left(\frac{4}{5}\right)^{3} $
(ii) $\left[\left(\frac{3}{7}\right)^{-2}\right]^{-3}$
(iii) $\left(\frac{5}{9}\right)^{-2} \times\left(\frac{5}{3}\right)^{2} \div\left(\frac{1}{5}\right)^{-2} $
(iv) $2^{-1}\left[\left(\frac{5}{3}\right)^{4}+\left(\frac{3}{5}\right)^{-2}\right] \div \frac{17}{9}$
(v) $(-7)^{3} \times\left(\frac{1}{-7}\right)^{-4} \div(-7)^{10}$
Sol :
(i)
$\begin{aligned} 5^{3} \times\left(\frac{4}{5}\right)^{3} &=5^{3} \times \frac{4^{3}}{5^{3}} \\ &=4^{3} \times 5^{3-3} \\ &=4^{3} \\ &=\frac{1}{(4)^{-3}} \end{aligned}$
(ii)
$\begin{aligned}\left.\left(\frac{3}{7}\right)^{-2}\right]^{-3} &=\left(\frac{3}{7}\right)^{-2 x-3} \quad\left(\because\left(a^{m}\right)^{n}=a^{m \times n}\right) \\ &=\left(\frac{3}{7}\right)^{6} \\ &=\left(\frac{7}{3}\right)^{-6} \end{aligned}$
(iii) $\begin{aligned}\left(\frac{5}{9}\right)^{-2} \times\left(\frac{5}{3}\right)^{2} \div\left(\frac{1}{5}\right)^{-2} &=\frac{\left(\frac{9}{5}\right)^{2} \times\left(\frac{5}{3}\right)^{2}}{\left(\frac{1}{5}\right)^{-2}} \\ &=\frac{9^{2}}{5^{2}} \times \frac{5^{2}}{3^{2}} \times \frac{1^{2}}{5^{2}} \\ &=3^{4-2} \times 5^{2-4} \\ &=3^{+2} \times 5^{-2}=5^{-2} \times \frac{1}{(3)^{-2}} \\ &=\left(\frac{5}{3}\right)^{-2} \end{aligned}$
(iv) $2^{-1}\left[\left(\frac{5}{3}\right)^{4}+\left(\frac{3}{5}\right)^{-2}\right] \div \frac{17}{9}$
Sol :
=$\frac{1}{2}\left[\left(\frac{5}{3}\right)^{4}+\left(\frac{5}{3}\right)^{2}\right] \times \frac{9}{17}$
=$\frac{1}{2}\left[\left(\frac{5}{3}\right)^{2}\left[\left(\frac{5}{3}\right)^{2}+1\right]\right] \times \frac{9}{17}$
$=\frac{1}{2}\left[\frac{25}{9}\left(\frac{25}{9}+1\right)\right] \times \frac{9}{17}$
$=\frac{1}{2}\left[\frac{25}{9}\left(\frac{25+9}{9}\right)\right] \times \frac{9}{17}$
$=\frac{1}{2}\left[\frac{25}{9} \times \frac{34}{9}\right] \times \frac{9}{-17}$
$=\frac{1}{2}\left[\frac{25}{9} \times \frac{17 \times 2}{9} \times \frac{9}{17}\right]$
$=\frac{5^{2}}{3^{2}}=\left(\frac{5}{3}\right)^{2}$
$=\left(\frac{3}{5}\right)^{-2}$
(v) $(-7)^{3} \times\left(\frac{1}{-7}\right)^{-9} \div(-7)^{10}$
Sol :
$=(-7)^{3} \times(-7)^{9} \div(-7)^{10}$
$=\frac{(-7)^{3} \times(-7)^{9}}{(-7)^{10}}$
$=(-7)^{3+9-10}$
$=(-7)^{2}=(7)^{2}$
$=\left(\frac{1}{-7}\right)^{-2}$
Q9 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Simplify :
(i) $\frac{49 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}}$
(ii) $\frac{9^{3} \times 27 \times t^{4}}{(3)^{2} \times(3)^{4} \times t^{2}}$
(iii) $\frac{\left(3^{-2}\right)^{2} \times\left(5^{2}\right)^{-3} \times\left(t^{-3}\right)^{2}}{\left(3^{-2}\right)^{5} \times\left(5^{3}\right)^{-2} \times\left(t^{-4}\right)^{3}}$
(iv) $ \frac{2^{-5} \times 15^{-5} \times 500}{5^{-6} \times 6^{-5}}$
Sol :
(i)
$\begin{aligned} \frac{49 \times z^{-3}}{7^{-3} \times 10 \times z^{-5}} &=\frac{7^{2} \times z^{-3}}{7^{-3} \times 10 \times z^{-5}} \\ &=\frac{7^{2+3} \times z^{-3+5}}{10} \quad\left(\because \frac{a^{n}}{a^{n}}=a^{m-n}\right) \\ &=\frac{7^{5} \times z^{2}}{10} . \end{aligned}$
(ii)
$\begin{aligned} \frac{9^{3} \times 27 \times t^{4}}{(3)^{2} \times(3)^{4} \times t^{2}} &=\frac{\left(3^{2}\right)^{3} \times 3^{3} \times t^{4}}{(3)^{2} \times(3)^{4} \times t^{2}} \\ &=3^{6} \times 3^{3} \times t^{4} \times 3^{-2} \times 3^{-4} \times t^{-2} \\ &=3^{6+3-2-4} \times t^{4-2}\left(\because a^{m} \times a^{n}=a^{m+n}\right) \\ &=3^{3} \times t^{2} \\ &=27 \times t^{2} \end{aligned}$
(iii) $\frac{\left(3^{-2}\right)^{2} \times\left(5^{2}\right)^{-3} \times\left(t^{-3}\right)^{2}}{\left(3^{-2}\right)^{5} \times\left(5^{3}\right)^{-2} \times\left(t^{-4}\right)^{3}}=\frac{3^{-4} \times 5^{-6} \times t^{-6}}{3^{-10} \times 5^{-6} \times t^{-12}}\left(\therefore\left(a^{n}\right)^{n}=a^{m \times n}\right)$
$\begin{aligned} &=3^{-4+10} \times 5^{-6+6} \times t^{-6+12}\left(\because \frac{a^{m}}{a^{n}}=a^{m-n}\right) \\ &=3^{6} \times 5^{0} \times t^{6} \\ &=3^{6} \times t^{6} \quad\left(\therefore a^{0}=1\right) \end{aligned}$
(iv)
$\begin{aligned} \frac{2^{-5} \times 15^{-5} \times 500}{5^{-6} \times 6^{-5}} &=\frac{2^{-5} \times(3 \times 5)^{-5} \times 5^{3} \times 2^{2}}{5^{-6} \times(2 \times 3)^{-5}} \\ &=\frac{2^{-5+2} \times 3^{-5} \times 5^{-5} \times 5^{3}}{5^{-6} \times 2^{-5} \times 3^{-5}} \\ &=2^{-5+2+5} \times 3^{-5+0+5} \times 5^{-5+6+3}=2^{2} \times 3 \times 5^{4} \\ &=2500 . \end{aligned}$
Q10 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
By what number should $\left(\frac{3}{-2}\right)^{-3}$ be divided to get $\left(\frac{2}{3}\right)^2$
Sol :
Let the number which should divide $\left(\frac{3}{-2}\right)^{-3}$ to got
$\left(\frac{2}{3}\right)^{2}$ be $x .$
$\Rightarrow \quad \frac{\left(\frac{3}{-2}\right)^{-3}}{x}=\left(\frac{2}{3}\right)^{2}$
$\Rightarrow \frac{3^{-3}}{(-2)^{-3}}=\frac{x \times 2^{2}}{3^{2}}$
Cross multiply
$\Rightarrow \quad 3^{-3} \times 3^{2}=x \times 2^{2} \times(-2)^{-3}$
$\Rightarrow \quad 3^{-3+2}=x \times(-2)^{2-3} \quad 3 \quad\left(\therefore(-a)^{2}=(a)^{2}\right)$
$\Rightarrow \quad 3^{-1}=x \times(-2)^{-1}$
$\begin{aligned} x &=\frac{3^{-1}}{(-2)^{-1}} \\ \therefore \quad x &=-2 / 3 \end{aligned}$
Q11 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Find the value of m for which 9m ÷ 3-2 = 94.
Sol :
$\begin{aligned} & 9^{m} \div 3^{-2}=9^{4} . \\ \Rightarrow & \frac{9^{m}}{3^{-2}}=9^{4} \\ \Rightarrow & 9^{m} \times 9=9^{4} \quad\left(\because \frac{1}{a^{m}}=\right.\\ \Rightarrow & 9^{m+1}=9^{4} \\ & \text { If } \quad a^{m}=a^{n} \quad \text { then } \quad m=n \\ \Rightarrow & m+1=4 \end{aligned}$
$\therefore m=3$
Q12 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Question 12
If $\left(\frac{-5}{7}\right)^{-4} \times\left(\frac{-5}{7}\right)^{12}=\left\{\left(\left(\frac{-5}{7}\right)^{3}\right)\right]^{x} \times\left(\frac{-5}{7}\right)^{-1}$
Find the value of x
Sol :
$\left(\frac{-5}{7}\right)^{-4} \times\left(\frac{-5}{7}\right)^{12}=\left\{\left(\left(\frac{-5}{7}\right)^{3}\right)\right]^{x} \times\left(\frac{-5}{7}\right)^{-1}$
$\left(\frac{-5}{7}\right)^{-4+12}=\left(\frac{-5}{7}\right)^{3 x-1}$
$\Rightarrow \quad-4+12=3 x-1 \quad\left(\because a^{m}=a^{n} \Rightarrow m=n\right)$
$\Rightarrow \quad 3 x=9$
$\therefore x=3$
Q13 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
Find x if $\left(\frac{-2}{3}\right)^{-13} \times\left(\frac{3}{-2}\right)^{8}=\left(\frac{-2}{3}\right)^{-2 x+1}$
Sol :
$\left(\frac{-2}{3}\right)^{-13} \times\left(\frac{3}{-2}\right)^{8}=\left(\frac{-2}{3}\right)^{-2 x+1}$
$\Rightarrow\left(\frac{-2}{3}\right)^{-13} \times\left(\frac{-2}{3}\right)^{-8}=\left(\frac{-2}{3}\right)^{-2 x+1} \quad\left(\because\left(\frac{p}{q}\right)^{m}=\left(\frac{q}{p}\right)^{-m}\right)$
$\Rightarrow\left(\frac{-2}{3}\right)^{-13-8}=\left(\frac{-2}{3}\right)^{-2 x+1}$
$\begin{aligned} \therefore-21=-2 x+1 &\left(\because a^{m}=a^{n} \text { then } m=n\right) \\ \Rightarrow &-2 x=-22 \\ \therefore x &=11 \end{aligned}$
Q14 | Ex-2.1 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
(i) If $5^{2x-1}=\frac{1}{(125)^{x-3}}$, find x
(ii) If $\frac{9^{n} \times 3^{5} \times(27)^{3}}{3 \times(81)^{4}}=27$ , find x
Sol :
(i) $5^{2 x-1}=\frac{1}{(125)^{x-3}} \Rightarrow 5^{2 x-1}=\frac{1}{\left(5^{3}\right)^{x-3}}$
$\Rightarrow \quad 5^{2 x-1} \times 5^{3(x-3)}=1$
$\Rightarrow \quad 5^{(2 x-1)+3(x-3)}=5 \quad\left(\because a^{0}=1\right)$
$\therefore \quad 2 x-1+3 x-9=0$
$\Rightarrow 5 x=10$
$\therefore x=2$
(ii) $\frac{9^{n} \times 3^{5} \times(27)^{3}}{3 \times(81)^{4}}=27$
$\Rightarrow \frac{\left(3^{2}\right)^{n} \times 3^{5} \times\left(3^{3}\right)^{3}}{3 \times\left(3^{4}\right)^{4}}=3^{3}$
$\Rightarrow 3^{(2 n+5+9)(-1-16)} \times 3^{3}$
$\Rightarrow 3^{2 n-3}=3^{3}$
$\therefore \quad 2 n-3=3$
$\Rightarrow n=3$
