Question

Class 9 Please

$\: if \: \: \dfrac{ {9}^{n} \times {3}^{2} \times {3}^{n} - {27}^{n} }{{3}^{3m} \times {2}^{3} } = \dfrac{1}{27}$
Prove that m - n = 1

It is from RD Sharma please guys... ​

Answer 1

Solution :

We have,

$\star \: \sf \: \dfrac{ {9}^{n } \times {3}^{2} \times {3}^{n} - {27}^{n} }{ {3}^{3m} \times {2}^{3} } = \dfrac{1}{27}$

$\longrightarrow \sf \: \dfrac{ {3}^{2n} \times {3}^{2} \times {3}^{n} - {3}^{3n} }{ {3}^{3m} \times {2}^{3} } = \dfrac{1}{27}$

$\longrightarrow \sf \: \dfrac{ {3}^{2n + 2 + n} - {3}^{3n} }{{3}^{3m} \times {2}^{3} } = \dfrac{1}{ {3}^{3} }$

$\longrightarrow \sf \: \dfrac{ {3}^{3n + 2} - {3}^{3n} }{ {3}^{3m} \times {2}^{3} } = \dfrac{1}{ {3}^{3} }$

$\longrightarrow \sf \dfrac{ {3}^{3n} \times {3}^{2} - {3}^{3n} }{ {3}^{3m} \times {2}^{3} } = \dfrac{1}{ {3}^{3} }$

$\longrightarrow \sf \dfrac{ {3}^{3n} ( {3}^{2} - 1) }{ {3}^{3m} \times 8} = \dfrac{1}{ {3}^{3} }$

$\longrightarrow \sf \dfrac{ {3}^{3n} \times 8}{ {3}^{3m} \times 8} = \dfrac{1}{ {3}^{3} }$

$\longrightarrow \sf {3}^{3n - 3m} = {3}^{ - 3}$

Bases are equal power will also be equal

$\longrightarrow \sf 3n - 3m = - 3$

$\longrightarrow \sf n - m = - 1$

$\longrightarrow \sf m - n = 1$

Hence, Proved!

$\dag \: \large \boxed{ \red{ \sf \: You \: must \: know : }}$

If a,b are positive real numbers and m,n are rational numbers,then :

$\sf \: i) \: \large {a}^{m} \times {a}^{n} = {a}^{m + n}$

$\sf \: ii) \large \: {a}^{m} \div {a}^{n} = {a}^{m - n}$

$\sf \: iii) \large \: ( {a}^{m} ) ^{n} = {a}^{mn}$

$\sf \: iv) \large \: {a}^{ - n} = \dfrac{1}{ {a}^{n} }$

$\sf \: v) \large \: ( {ab})^{n} = {a}^{n} {b}^{n}$

$\sf \: vi) \large \: ( \dfrac{a}{b} ) ^{n} = \dfrac{ {a}^{n} }{ {b}^{n} }$

Answer 2

Answer:

$\underline{\bigstar\:\textsf{According to the Question :}}$

$:\implies\tt \dfrac{9^{n } \times {3}^{2} \times {3}^{n} - {27}^{n} }{ {3}^{3m} \times {2}^{3} } = \dfrac{1}{27}\\\\\\:\implies\tt \dfrac{ {3}^{2n} \times {3}^{2} \times {3}^{n} - {3}^{3n} }{ {3}^{3m} \times {2}^{3} } = \dfrac{1}{27}\\\\\\:\implies\tt \dfrac{ {3}^{2n + 2 + n} - {3}^{3n} }{{3}^{3m} \times {2}^{3} } = \dfrac{1}{ {3}^{3} }\\\\\\:\implies\tt \dfrac{ {3}^{3n + 2} - {3}^{3n} }{ {3}^{3m} \times {2}^{3} } = \dfrac{1}{3^3}\\\\\\:\implies\tt \dfrac{3^{3n} \times {3}^{2}- {3}^{3n} }{ {3}^{3m} \times {2}^{3} } = \dfrac{1}{3^3}\\\\\\:\implies\tt \dfrac{3^{3n} ( {3}^{2} - 1) }{3^{3m} \times 8} = \dfrac{1}{3^3}\\\\\\:\implies\tt \dfrac{3^{3n} \times 8}{ {3}^{3m} \times 8} = \dfrac{1}{3^3}\\\\\\:\implies\tt {3}^{3n - 3m} = {3}^{-3}$

${\scriptsize\qquad\bf{\dag}\:\:\textsf{Base is Equal, So Power will be Equal}}\\\\:\implies\tt 3n - 3m = -\:3\\\\\\:\implies\tt 3(n-m)=-\:3\\\\\\:\implies\tt (n-m)=\dfrac{-\:3}{3}\\\\\\:\implies\tt (n-m)=-\:1\\\\\\:\implies\underline{\boxed{\red{\tt (m-n)=1}}}$

$\rule{200}{2}$

$\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}$