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# Solution for Show that Log X X Has a Maximum Value at X = E ? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Show that $\frac{\log x}{x}$ has a maximum value at x = e ?

#### Solution

$\text { Here },$

$f\left( x \right) = \frac{\log x}{x}$

$\Rightarrow f'\left( x \right) = \frac{1 - \log x}{x^2}$

$\text { For the local maxima or minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow \frac{1 - \log x}{x^2} = 0$

$\Rightarrow 1 = \log x$

$\Rightarrow \log e = \log x$

$\Rightarrow x = e$

$\text { Now,}$

$f''\left( x \right) = \frac{x^2 \left( \frac{- 1}{x} \right) - 2x\left( 1 - \log x \right)}{x^4} = \frac{- 3 + 2 \log x}{x^3}$

$\Rightarrow f''\left( e \right) = \frac{- 3 + 2 \log e}{e^3} = \frac{- 1}{e^3} < 0$

$\text { So, x = e is the point of local maximum }.$

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Solution Show that Log X X Has a Maximum Value at X = E ? Concept: Graph of Maxima and Minima.
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