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

4

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

8

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

Open in App

Solution

The correct option is B $2$
Given that:
${log}_{x} a + {log}_{a} x = \frac{log a}{log x} + \frac{log x}{log a} ∵ A . M \geq G . M ∴ \frac{1}{2} (\frac{log a}{log x} + \frac{log x}{log a}) \geq \sqrt{\frac{log a}{log x} \frac{log x}{log a}} \Rightarrow {log}_{x} a + {log}_{a} x \geq 2$

Suggest Corrections

Similar questions

x

\sqrt{{log}_{x} (a x)^{1 / 5} + {log}_{a} (a x)^{1 / 5}} + \sqrt{{log}_{a} {(\frac{x}{a})}^{1 / 5} + {log}_{x} {(\frac{a}{x})}^{1 / 5}} = a

\frac{d y}{d x}

y = l o g_{a} x + l o g_{x} a + l o g_{x} x + l o g_{a} a

y = l o g_{a} x + l o g_{x} a + l o g_{x} x + l o g_{a} a

\frac{d y}{d x} =

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

y + l o g_{a} x + l o g_{x} a + l o g_{a} a

\frac{d y}{d x} =

Join BYJU'S Learning Program