The greatest value of $(4 {log}_{10} x - {log}_{x} (0.0001))$ for $0 < x < 1$ is

4

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

- 4

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

8

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

- 8

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

Open in App

Solution

The correct option is A $8$
$(4 l o g_{10} x - l o g_{x} (0.0001)$
$(4 l o g_{10} x - l o g_{x} \frac{1}{10^{4}}$
$(4 l o g_{10} x - (l o g_{x} 1 - l o g_{x} 10^{4})$
$(4 l o g_{10} x - 0 + l o g_{x} 10^{4})$
$(4 l o g_{10} x + l o g_{x} 10^{4})$
$4 (l o g_{10} x + l o g_{x} 10)$
$4 (l o g_{10} x + \frac{1}{l o g_{10} x})$
$A M \geq G M$
$\frac{x + y}{2} \geq \sqrt{x y}$
$l o g_{10} x + \frac{1}{l o g_{10} x} \geq 2$
$4 (l o g_{10} x + \frac{1}{l o g_{10} x}) \geq 8$

Suggest Corrections

Similar questions

(4 {log}_{10} x - {log}_{x} (.0001)) f o r 0 ⟨ x ⟨ 1

1 + 3 {log}_{10} \sqrt{2 + x} + 4 {log}_{10} \sqrt{2 - x} = 3 {log}_{10} \sqrt{4 - x^{2}}

x

1 + 3 {log}_{10} \sqrt{2 + x} + 4 {log}_{10} \sqrt{2 - x} = 3 {log}_{10} \sqrt{4 - x^{2}}

x

For what value of $k, (4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0$ , is a perfect square.

λ

2 x^{2} + (λ - 1) x + 8 = 0

x^{2} - 8 x + λ + 4 = 0

Join BYJU'S Learning Program