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answers:
Elizabeth 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
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JG say: I am a little confused by the notation. Did you mean (1/27)^(3n+1) = √3 / 81 ? If so, you can rewrite the equation:
(3^(-3))^(3n+1) = 3^(1/2)/(3^(4))
3^(-9n-3) = 3^(-7/2)
-9n-3 = -7/2
-9n = 3-7/2 = -1/2
n = 1/18
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Philip say: (1/27)^(3n +1) = (1/81)rt3, ie., (1/27)^(3n) = 3^(-1/2). Take ln of both sides
getting -3n*ln(27) = -(1/2)*ln(3), ie., 9n*ln(3) = (1/2)*ln(3), ie., n = (1/18).
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Krishnamurthy say: (1/27)^3n + 1 = √3 / 81
3^3n = √3 - 81
√3^6n = √3^(1 - 27√3)
6n = 1 - 27√3
Solution:
n ≈ -7.6276
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sepia say: (1 / 27)^(3n + 1) = √3 / 81
Solution:
n = 1/18
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Ash say: (1/27)^(3n+1) = (√3)/81
(1/27)^(3n) * (1/27)¹ = (√3)/81
(1/3³)^(3n) = (27)(√3)/81
(3⁻³)^(3n) = (√3)/3