Math, asked by Lahariyenishetti, 3 months ago

81^-2÷729^-1=9^2x solve x ​

Answers

Answered by fasmohammad3500
2

Answer:

7

Step-by-step explanation:

Given,

81^-2÷729^1-x = 9^2x

(9^2)^-2÷(9^3)^(1-x) = 9^2x

9^-4÷9^(3-3x) = 9^2x

9^-4 = 9^2x × 9^(3-3x)

9^-4 = 9^(2x+3-3x)

For the power x --

-4 = 2x+3-3x

-4 = x+3

x = 3+4

Answered by pro2007
0

Answer:

Valueofx=5

Step-by-step explanation:

\begin{gathered}Given \\\frac{(81)^{-1}}{(729)^{1-x}}=9^{2x}\end{gathered}

Given

(729)

1−x

(81)

−1

=9

2x

\implies \frac{(9^{2})^{-1}}{(9^{3})^{1-x}}=9^{2x}⟹

(9

3

)

1−x

(9

2

)

−1

=9

2x

\implies \frac{9^{2\times (-1)}}{9^{3\times (1-x)}}=9^{2x}⟹

9

3×(1−x)

9

2×(−1)

=9

2x

\begin{gathered} By \: Exponential \:Law,\\(a^{m})^{n}=a^{mn}\end{gathered}

ByExponentialLaw,

(a

m

)

n

=a

mn

\implies \frac{9^{-2}}{9^{3-3x}}=9^{2x}⟹

9

3−3x

9

−2

=9

2x

\implies 9^{-2-(3-3x)}=9^{2x}⟹9

−2−(3−3x)

=9

2x

\begin{gathered} By \: Exponential \:Law,\\\frac{a^{m}}{a^{n}}=a^{m-n}\end{gathered}

ByExponentialLaw,

a

n

a

m

=a

m−n

\implies 9^{-2-3+3x}=9^{2x}⟹9

−2−3+3x

=9

2x

\implies 9^{-5+3x}=9^{2x}⟹9

−5+3x

=9

2x

\implies -5+3x=2x⟹−5+3x=2x

\begin{gathered} By \: Exponential \:Law,\\If\:a^{m}=a^{n}\: then \: m=n\end{gathered}

ByExponentialLaw,

Ifa

m

=a

n

thenm=n

\implies 3x-2x=5⟹3x−2x=5

\implies x = 5⟹x=5

Therefore,

Value \: of \:x = 5Valueofx=5

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