Question

solve for x:81^-2 divide 729^1-x=9^2x

Answer 1

Answer:

$Value \: of \:x = 5$

Step-by-step explanation:

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

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

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

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

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

$\implies 9^{-2-(3-3x)}=9^{2x}$

$By \: Exponential \:Law,\\\frac{a^{m}}{a^{n}}=a^{m-n}$

$\implies 9^{-2-3+3x}=9^{2x}$

$\implies 9^{-5+3x}=9^{2x}$

$\implies -5+3x=2x$

$By \: Exponential \:Law,\\If\:a^{m}=a^{n}\: then \: m=n$

$\implies 3x-2x=5$

$\implies x = 5$

Therefore,

$Value \: of \:x = 5$

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Answer 2

Answer is 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

x = 7