If ${log}_{3} [{log}_{4} ({log}_{2} x)] = 0$ then the value of $x$ is

16

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!

64

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

32

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

Open in App

Solution

The correct option is A $16$
${log}_{3} [{log}_{4} ({log}_{2} x)] = 0$
$\Rightarrow {log}_{4} ({log}_{2} x) = 3^{0} = 1$
$\Rightarrow {log}_{2} x = 4^{1}$
$\Rightarrow x = 2^{4}$
$\Rightarrow x = 16$

Suggest Corrections

Similar questions

{lim}_{x \to 0} \frac{log (3 + x) - log (3 - x)}{x} = k

k

α, β

x^{2} - 2 x + 4 = 0

α^{3}, β^{3}

8163264

log \frac{9}{8} - log \frac{27}{32} + log \frac{3}{4}

log \frac{9}{8} - log \frac{27}{32} + log \frac{3}{4}

Join BYJU'S Learning Program