Find the value of x in the following expression: 81⁻² ÷ 729¹⁻ˣ = 9²ˣ
Answers
Answer:
Given(729)1−x(81)−1=92x
\implies \frac{(9^{2})^{-1}}{(9^{3})^{1-x}}=9^{2x}⟹(93)1−x(92)−1=92x
\implies \frac{9^{2\times (-1)}}{9^{3\times (1-x)}}=9^{2x}⟹93×(1−x)92×(−1)=92x
\begin{gathered} By \: Exponential \:Law,\\(a^{m})^{n}=a^{mn}\end{gathered}ByExponentialLaw,(am)n=amn
\implies \frac{9^{-2}}{9^{3-3x}}=9^{2x}⟹93−3x9−2=92x
\implies 9^{-2-(3-3x)}=9^{2x}⟹9−2−(3−3x)=92x
\begin{gathered} By \: Exponential \:Law,\\\frac{a^{m}}{a^{n}}=a^{m-n}\end{gathered}ByExponentialLaw,anam=am−n
\implies \frac{9^{-2}}{9^{3-3x}}=9^{2x}⟹93−3x9−2=92x
\implies 9^{-2-(3-3x)}=9^{2x}⟹9−2−(3−3x)=92x
\begin{gathered} By \: Exponential \:Law,\\\frac{a^{m}}{a^{n}}=a^{m-n}\end{gathered}ByExponentialLaw,anam=am−n
\implies 9^{-2-3+3x}=9^{2x}⟹9−2−3+3x=92x
\implies 9^{-5+3x}=9^{2x}⟹9−5+3x=92x
\implies -5+3x=2x⟹−5+3x=2x
\begin{gathered} By \: Exponential \:Law,\\If\:a^{m}=a^{n}\: then \: m=n\end{gathered}ByExponentialLaw,Ifam=anthenm=n
\implies 3x-2x=5⟹3x−2x=5
\implies x = 5⟹x=5
Therefore,
Value \: of \:x = 5Valueofx=5