Question

Solve the following: (81)^-4 ÷ (729)^2-x = 9^4x

Answer 1

Answer:

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Answer 2

Answer:

$81^{-4}$÷$729^{2-x}$=$9^{4x}$(Given)

Step-by-step explanation:

So,81 = $3^{4}$, 729= $3^{6}$ , 9 = $3^{2}$

   $(3^{4})^{-4}$ ÷  $(3^{6})^{2-x}$ =  $(3^{2})^{4x}$

  as the rule in exponents , $(a^{n})^{m} = a^{n *m}$  (* means multiplication [×])

$3^{4*-4}$ ÷ $3^{6*(2-x)}$ = $3^{2*4x}$

$3^{-16}$ ÷ $3^{12-6x}$ = $3^{8x}$

another rule , that $a^{n}$ ÷ $a^{m} = a^{n-m}$

Hence,

$3^{-16-(12 - 6x)} = 3^{8x}$

$3^{-16+6x-12} = 3^{8x}$      ( as -(12-6x) = -12+6x = 6x + (-12)  )

$3^{6x-28} = 3^ {8x}$

There is a rule which states that when same bases (3) are in a equation(LHS and RHS) , then there powers are equated

Hence

6x - 28 = 8x

6x = 8x + 28

$\frac{6x}{8x} \\\\= \frac{3}{4}$