For $x > 1$ , the least value of the expression $2 {log}_{10} x - {log}_{x} (0 \cdot 01)$ is :

10

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

- 0 \cdot 01

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

2

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

None of these

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

Open in App

Solution

The correct option is D None of these
$x > 1$
$2 l o g_{10} x - l o g_{x} (0.01) = 2 l o g_{10} x - 2 l o g_{x} (0.1)$
$= 2 l o g_{10} x + 2 l o g_{x} 10$
$= 2 (l o g_{10} x + \frac{1}{l o g_{10} x})$ .......(1)
applying property of A.M. and G.M.
$l o g_{10} x + \frac{1}{l o g_{10} x} \geq 2$

From Eq (1) given expression $\geq 2 \times 2 \geq 4$

Suggest Corrections

Similar questions

2 {log}_{10} x - {log}_{x} (0.01)

x > 1

x > 1

2 {log}_{10} x - {log}_{x} 0.01

2 {log}_{10} x - {log}_{x} (0.01)

x > 1

2 l o g_{10} x - l o g_{x} 0.01

x > 0, x \neq 1.

\int_{0}^{1} f (x) d x = 1, \int_{0}^{1} x f (x) d x = a, \int_{0}^{1} x^{2} f (x) d x = a^{2}, then \int_{0}^{1} {(a - x)}^{2} f (x) d x

Join BYJU'S Learning Program