Find the value of 'x' if(12logx4)⋅(12logx16)⋅(12logx256)…=2

(12logx4)(12logx16)(12logx16)=22−logx4×2−logx16×2−logx256=22(−logx4+logx16+logx256+…)=2logx(4×16×256×⋯=1x22×24×28×…=1x22+4+8+…=1xNow say we have, 2+4+8+…; this can be written as 2+4+8+…+2n when n→∞.2+4+8+…+2n=2(2n−1)x=limn→∞(122(2n−1))

Answered: "Find the value of 'x' if(12logx4)⋅(12logx16)⋅(12logx256)…=2"

High School
Calculus and Analysis
Calculus 2
Series

Aileen Rogers

Answered question

2022-04-06

Find the value of 'x' if

(\frac{1}{2^{\log_{x} 4}}) \cdot (\frac{1}{2^{\log_{x} 16}}) \cdot (\frac{1}{2^{\log_{x} 256}}) \dots = 2

Answer & Explanation

cutimnm135imsa

Beginner2022-04-07Added 21 answers

(\frac{1}{2^{\log_{x} 4}}) (\frac{1}{2^{\log_{x} 16}}) (\frac{1}{2^{\log_{x} 16}}) = 2

2^{- \log_{x} 4} \times 2^{- \log_{x} 16} \times 2^{- \log_{x} 256} = 2

2^{(- \log_{x} 4 + \log_{x} 16 + \log_{x} 256 + \dots)} = 2

\log_{x} (4 \times 16 \times 256 \times \dots = \frac{1}{x}

2^{2} \times 2^{4} \times 2^{8} \times \dots = \frac{1}{x}

2^{2 + 4 + 8 + \dots} = \frac{1}{x}

Now say we have,

2 + 4 + 8 + \dots

; this can be written as

2 + 4 + 8 + \dots + 2^{n}

when

n \to \infty .2 + 4 + 8 + \dots + 2^{n} = 2 (2^{n} - 1)

x = lim_{n \to \infty} (\frac{1}{2^{2 (2^{n} - 1)}})

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