You visited us 2 times! Enjoying our articles? Unlock Full Access!

Convexity

For 0<a<x, ...

Question

For $0 < a < x$ , the minimum value of the function ${log}_{x} a + {log}_{a} x$ is:

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $2$
$y = l o g_{x} a + l o g_{a} x$
$= \frac{l n a}{l n x} + \frac{l n x}{l n a}$
$= \frac{l n a}{l n x} + \frac{1}{\frac{l n a}{l n x}}$
$= l o g_{x} a + \frac{1}{l o g_{x} a}$
Applying $A . M \geq G . M$
Hence
$\frac{l o g_{x} a + \frac{1}{l o g_{x} a}}{2} \geq \sqrt{l o g_{x} a . \frac{1}{l o g_{x} a}}$
$l o g_{x} a + \frac{1}{l o g_{x} a} \geq 2$
$y \geq 2$ .
Hence the minimum value of the function is 2.

Suggest Corrections

Similar questions

\frac{log x}{x} \in 0 < x < \infty

(a > 1, x > 1)

(0 < a < 1, 0 < x < 1)

{log}_{a} x > 0

{log}_{a} x

(0 < a < 1, x > 1)

(a > 1, 0 < x < 1)

{log}_{a} x < 0

a > b

{log}_{b} a > 1

{log}_{a} b < 1.

\frac{{log}_{a} (3.162)}{{log}_{a} (2 / 3)} .

f (x) = {log}_{a} x

0 < a < 1

x > 0

lim x \to \infty \frac{{cot}^{- 1} (x^{- a} {log}_{a} x)}{{sec}^{- 1} (a^{x} {log}_{x} a)}, (a > 1)

f (x) = 2 x^{3} - 3 x^{2_{-}} 12 x + 8

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 8, f^{'} (2) = 0

f^{''} (2) > 0

Join BYJU'S Learning Program