If log x - log 2 - 12log x- then

The correct option is B x ≥ 4 logx ≥ log2+1/2logx Which implies 1/2logx ≥ log2 Which implies logx ≥ log4 Therefore , we get x ≥ 4 ...

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Algebra of Derivatives

If log x ≥ ...

Question

If $l o g x \geq l o g 2 + \frac{1}{2} l o g x$ , then

x \geq 2

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x \leq 2

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x \leq 4

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x \geq 4

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x \geq 1

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l o g x \geq l o g 2 + \frac{1}{2} l o g x

2 log (x + 1) - log (x^{2} - 1) = log 2,

x =

\frac{1}{2} log x + \frac{1}{2} log y + log 2 = log (x + y)

x

y

log (\frac{x + y}{5}) = \frac{1}{2} (log x + log y)

\frac{x}{y} + \frac{y}{x} =

{x \in N : 2 + {log}_{2} \sqrt{x + 1} > 1 - {log}_{1 / 2} \sqrt{4 - x^{2}}}

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