Question

please tell answer of this question
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Answer 1

GIVEN :-

$\implies \bf \: 9 ^{x} \times 3 ^{2} \times \bigg \{ {3}^{ \frac{ - x}{2} } \bigg \} ^{ - 2} = \dfrac{1}{27}$

TO FIND :-

$\implies \bf \: The \: value \: of \: x$

SOLUTION :-

$\to \rm \: 9 ^{x} \times 3 ^{2} \times \bigg \{ {3}^{ \frac{ - x}{2} } \bigg \} ^{ - 2} = \dfrac{1}{27} \\ \\ \to \: \rm \: 3 ^{2x} \times 3 ^{2} \times \bigg \{3 ^{x} \bigg \} \: = \dfrac{1}{27} \\ \\ \star \: \bf \: By \: using \: \: x ^{a} \times x ^{b} = {x}^{a \: + b} \\ \\ \to \rm \: (3) ^{2x + 2 + x} = \dfrac{1}{27} \\ \\ \to \rm \: (3) ^{3x + 2} = (3) ^{ - 3}$

➠ Here the bases of L.H.S and R.H.S are same, so there exponents will be also same.

Hence by comparing both the sides we get,

$\to \rm \: 3x + 2 = - 3 \\ \\ \to \rm3x = - 3 - 2 \\ \\ \to \rm \: 3x = - 5 \\ \\ \to \rm \: \boxed{ \red{x = \dfrac{ - 5}{3} }}$

Hence the value of x is -5/3.

ADDITIONAL INFORMATION :-

$\boxed{\begin{minipage}{7cm} \\ \sf{ \implies \bf \sqrt[n]{ \sqrt[m]{ \sqrt[p]{((a^{x} )^{y}) ^{z} } } } = (a ^{xyz} )^{ \frac{1}{mnp} } = a ^{ \frac{xyz}{mnp} } } \\ \\ \sf{ \implies a^m \times a^n = a^{m+n} } \\ \\ \sf{ \implies {a}^{m} \times b^m = ab^m } \\ \\ \sf{ \implies \dfrac{a^m}{a^n} = a^{m - n} ( \tt{ If \: m > n} ) } \\ \\ \sf{ \implies \dfrac{a^m}{ a^n} = \dfrac{ 1}{ a^{n-m} } ( \tt{ If \: n > m )} } \\ \\ \sf{ \implies (a^m)^n = a^{mn} } \\ \\ \sf{ \implies a^{-n} = \dfrac{1}{ a^n} }\end{minipage}}$