Very Hard Maths Question!!???! (1 / 27)^3n+1 = √3 / 81 Find N :( Thanks?
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Very Hard Maths Question!!???! (1 / 27)^3n+1 = √3 / 81 Find N :( Thanks?

[From: ] [author: ] [Date: 17-04-09] [Hit: ]
and : n = 1 / 18-katie say: i got -19683+343√3 does this sound right?-Steve A say: (1/27)^(3n+1) = sqrt(3) /81 (3^(-3))^(3n+1) = 3^(1/2)3^(-4) 3^(-9n +3) = 3^(-3.5) -9n + 3 = -3.5 6.5 = 9n 6.5/9 = n 13/18 = n-dylan say: Add brackets is it(1 / 27)^3*(n+1) (1 / 27)^3*n+1 (1 / 27)^(3n)+1 (1 / 27)^(3n+1)-Cecile say: i got -19683+343√3 does this sound right?......
3^(-9n) = 1/√3
3^(-9n) = 3^(-1/2)

-9n = -1/2
n = 1/18
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Huh!? say: (1 / 27)^(3n + 1) = √(3) / 81

(1 / 27)^(3n) * (1 / 27)^(1) = √(3) / 81
Multiply both sides of the equation by 27:

(1 / 27)^(3n) = √(3) * (27 / 81)

(1/ 27)^(3n) = √(3) / 3

Take √(3) / 3, and multiply by √(3) / √(3) which is equivalent to 1.

√(3) / 3 * [√(3) / √(3)]
= 3 / 3(√3)

= 1 / (√3), and that comes out to √(1 / 3) or (1 / 3)^(1 / 2) # This was hiding in the expression. The textbook writers decided to take the square root out of the denominator!

(1 / 27)^(3n) = (1 / 3)^(1 / 2)
Invert both sides of the equation: Raise to the negative 1th power.

[(1 / 27)^(3n)]^(-1) = [(1 / 3)^(1 / 2)]^(-1)

[(1 / 27)^(-1)]^(3n) = [(1 / 3)^(-1)]^(1/2)

(27 / 1)^(3n) = (3 / 1)^(1 / 2)

(3^3)^(3n) = 3^(1 / 2)

3^[(3 * 3n)] = 3^(1 / 2)
The bases of both exponents are the same so equate exponents:

9n = 1 / 2
Divide both sides of the equation by 9, and :

n = 1 / 18
-
katie say: i got -19683+343√3 does this sound right?
-
Steve A say: (1/27)^(3n+1) = sqrt(3) /81
(3^(-3))^(3n+1) = 3^(1/2)3^(-4)
3^(-9n +3) = 3^(-3.5)
-9n + 3 = -3.5
6.5 = 9n
6.5/9 = n
13/18 = n
-
dylan say: Add brackets is it
(1 / 27)^3*(n+1)
(1 / 27)^3*n+1
(1 / 27)^(3n)+1
(1 / 27)^(3n+1)
-
Cecile say: i got -19683+343√3 does this sound right?
-

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