The solution set of 2 log 2log 2 x -log 1 - 2log 22 -2 x -1 isA- 2-4B- 4-8C- 1 - 2-4D- 8

The correct option is D 8 log_22log_2x+log_1/2log_2(2√(2x))=1 log_2log_2x^2+log_1/2log_2(8x)^1/2=1 log_2log_2x^2+log_1/21/2+log_1/2log_28x=1 log_2log_2x^2 ...

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Standard IX

Mathematics

Fundamental Laws of Logarithms

The solution ...

Question

The solution set of $2 l o g_{2} l o g_{2} x + l o g_{1 / 2} l o g_{2} (2 \sqrt{2 x}) = 1$ is

{8}

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{2,4}

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{4,8}

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{1/2,4}

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Solution

The correct option is A
{8}

$l o g_{2} 2 l o g_{2} x + l o g_{1 / 2} l o g_{2} (2 \sqrt{2 x}) = 1 l o g_{2} l o g_{2} x^{2} + l o g_{1 / 2} l o g_{2} (8 x)^{\frac{1}{2}} = 1 l o g_{2} l o g_{2} x^{2} + l o g_{1 / 2} \frac{1}{2} + l o g_{1 / 2} l o g_{2} 8 x = 1 l o g_{2} l o g_{2} x^{2} = - l o g_{1 / 2} l o g_{2} 8 x l o g_{2} l o g_{2} x^{2} = l o g_{2} l o g_{2} 8 x \Rightarrow x^{2} = 8 x \Rightarrow x^{2} - 8 x = 0 x (x - 8) = 0 x = 8$

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The solution set of $2 l o g_{2} l o g_{2} x + l o g_{1 / 2} l o g_{2} (2 \sqrt{2 x}) = 1$ is

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The solution set of 2log2log2x+log1/2log2(2√2x)=1 is

The correct option is A {8} log22log2x+log1/2log2(2√2x)=1log2log2x2+log1/2log2(8x)12=1log2log2x2+log1/212+log1/2log28x=1log2log2x2=−log1/2log28xlog2log2x2=log2log28x⇒x2=8x⇒x2−8x=0x(x−8)=0x=8

The solution set of $2 l o g_{2} l o g_{2} x + l o g_{1 / 2} l o g_{2} (2 \sqrt{2 x}) = 1$ is