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If f ( x )= _{x}1/9-_{3}x^{2} ( x > 1 ), then max f ( x )

Question

If $f (x) = lo g_{x} 1/9 - lo g_{3} x^{2} (x > 1)$ , then max $f (x)$ is equal to

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Benjamin Discussed

f (x) = lo g_{x} 1/9 - lo g_{3} x^{2} (x > 1)

, then max

f (x)

is equal to

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Let,

y = lo g_{x} 1/9 - lo g_{3} x^{2} (x > 1)

= lo g_{x} (3^{- 2}) - 2 lo g_{3} x = - 2 lo g_{x} 3 - 2 lo g_{3} x

= - 2 (lo g_{x} 3 + lo g_{3} x)

Now since

x > 1 \Rightarrow lo g_{3} x, lo g_{x} 3 > 0

Thus using,

A . M \geq G . M

\Rightarrow \frac{lo g _{3} x + lo g _{x} 3}{2} \geq lo g_{x} 3 lo g_{3} x = 1, [∵ lo g_{a} b = \frac{1}{lo g _{b} a}]

Hence

(- y)_{ma x} = 2 \times 2 = 2

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Question Text	If $f (x) = lo g_{x} 1/9 - lo g_{3} x^{2} (x > 1)$ , then max $f (x)$ is equal to
Answer Type	Text solution:1
Upvotes	150