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Algebra Level 3

Determine the sum of all real solutions ( x ) (x) to the following equation:

log ⁡ 4 + ( 2 x + 1 2 x ) log ⁡ 3 = log ⁡ ( 3 x + 27 ) \log{4}+\bigg(\dfrac{2x+1}{2x}\bigg) \log{3} = \log{(\sqrt[x]{3}+27})

0 3 4 \dfrac{3}{4} 1 4 \dfrac{1}{4} 1 2 \dfrac{1}{2}

Satvik Golechha by Satvik Golechha , 21, India

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2 solutions

Chew-Seong Cheong
Dec 20, 2014

log ⁡ 4 + ( 2 x + 1 2 x ) log ⁡ 3 = log ⁡ ( 3 x + 27 ) \log{4}+\left( \dfrac {2x+1}{2x} \right) \log {3} = \log {(\sqrt [ x ]{ 3 } + 27)}

⇒ 4 ˙ 3 2 x + 1 2 x = 3 x + 27 ⇒ 4 ˙ 3 1 + 1 2 x = 3 1 x + 27 ⇒ 12 ˙ 3 1 2 x = 3 1 x + 27 \Rightarrow 4\dot{}3^{\frac {2x+1}{2x}} = \sqrt [ x ]{ 3 } + 27 \quad \Rightarrow 4\dot{} 3^{1+\frac {1}{2x} } = 3^{\frac {1}{x}} + 27 \quad \Rightarrow 12 \dot{} 3^{\frac {1}{2x} } = 3^{\frac {1}{x}} + 27

⇒ ( 3 1 2 x ) 2 − 12 ( 3 1 2 x ) + 27 = 0 ⇒ ( 3 1 2 x − 3 ) ( 3 1 2 x − 9 ) = 0 \Rightarrow (3^{\frac {1}{2x} })^2 - 12(3^{\frac {1}{2x}} ) + 27 = 0 \quad \Rightarrow (3^{\frac {1}{2x} } - 3) (3^{\frac {1}{2x} } - 9) = 0

⇒ { 3 1 2 x = 3 = 3 1 ⇒ 1 2 x = 1 ⇒ x = 1 2 3 1 2 x = 9 = 3 2 ⇒ 1 2 x = 2 ⇒ x = 1 4 \Rightarrow \begin {cases} 3^{\frac {1}{2x} } = 3 = 3^1 & \Rightarrow \dfrac {1}{2x} = 1 & \Rightarrow x = \frac {1}{2} \\ 3^{\frac {1}{2x} } = 9 = 3^2 & \Rightarrow \dfrac {1}{2x} = 2 & \Rightarrow x = \frac {1}{4} \end {cases}

The sum of all real roots of x x is 1 2 + 1 4 = 3 4 \frac {1}{2} + \frac {1}{4} = \boxed {\frac {3}{4}}

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answer is incorrect dear as n th root is defined for natural numbers n>=2 so x cannot be rational

hafiz khan - 5 years, 2 months ago

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Not really, it does not have to be restricted to only natural numbers. y x ≡ y 1 x \sqrt [x] y \equiv y ^{\frac{1}{x}} .

Chew-Seong Cheong - 5 years, 1 month ago

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yes dear this format is for natural number only you may check somewhere

hafiz khan - 5 years, 1 month ago
Ajit Athle
Dec 19, 2014

If 3^(1/x)=z then the given equation reduces to 12(z)^(1/2)=z+27 which yields z=9 or 81 giving x=1/2 or 1/4. Hence the sum = 3/4

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