Question

((729)^n/6×9^2n+1)÷(81^n×3^n-1)=?​

Answer 1

Answer:

the answer is 9 of your question hope it would be helpful

Answer 2

The answer is 27

Given problem:

$[ \frac{(729)^{\frac{n}{6}} . (9^{2n+1} )}{ ( 81^{n} . 3^{n-1} )} ]$  

Solution:

Take Numerator $(729)^{\frac{n}{6}} . (9^{2n+1} )$  

Here,  729  = 3⁶                          

⇒ 9²ⁿ⁺¹  = (3²)²ⁿ⁺¹ = 3⁴ⁿ⁺²            [ ∵ (aⁿ)ˣ = aⁿˣ ]  

⇒ $(729)^{\frac{n}{6}} . (9^{2n+1} )$  = $(3^{6} )^{\frac{n}{6}} . (3^{4n+2} )$ = $(3^{n} ). (3^{4n+2} )$  $= 3^{n+4n+2} = 3^{5n+2}$  

$(729)^{\frac{n}{6}} . (9^{2n+1} )$  = $3^{5n+2}$ _ (1)  

   

Take denominator $81^{n} . 3^{n-1}$  

⇒ 81ⁿ = (3⁴)ⁿ  = 3⁴ⁿ  

⇒ $81^{n} . 3^{n-1}$  = 3⁴ⁿ (3ⁿ⁻¹) =  3⁵ⁿ⁻¹

$81^{n} . 3^{n-1}$ =  3⁵ⁿ⁻¹ _ (2)

From (1) and (2)

$[ \frac{(729)^{\frac{n}{6}} . (9^{2n+1} )}{ ( 81^{n} . 3^{n-1} )} ]$ =  $\frac{3^{5n+2}}{3^{5n-1} }$ = $\frac{3^{5n} (3^{2}) }{(3^{5n}) (3^{-1}) }$ = 3²×3¹ = 3³ = 27      

 

Therefore, [(729)^n/6×9^2n+1] ÷ [ 81^n×3^n-1] = 27

#SPJ2