If $l o g_{2} (4^{x + 1} + 4) l o g_{2} (4^{x} + 1) = l o g_{2} 8,$ then x equals

-1

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

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

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

Open in App

Solution

The correct option is B
0

$l o g_{2} (4^{x + 1} + 4) l o g_{2} (4^{x} + 1) = l o g_{2} 8 l o g_{2} 4 (4^{x} + 1) l o g_{2} (4^{x} + 1) = 3 L e t 4^{x} + 1 = t l o g_{2} 4 t l o g_{2} t = 3 (l o g_{2} 4 + l o g_{2} t) l o g_{2} t = 3 (2 + l o g_{2} t) l o g_{2} t = 3 (l o g_{2} t)^{2} + 2 l o g_{2} t - 3 = 0 (l o g_{2} t)^{2} - l o g_{2} t + 3 l o g_{2} t - 3 = 0 l o g_{2} t (l o g_{2} t - 1) + 3 (l o g_{2} t - 1) = 0 l o g_{2} t = - 3 l o g_{2} t = 1 l o g_{2} (4^{x} + 1) = - 3 l o g_{2} (4^{x} + 1) = 1 4^{x} + 1 = \frac{1}{8} 4^{x} + 1 = 2 4^{x} \neq \frac{- 7}{8} 4^{x} = 1 x = 0$

Suggest Corrections

Similar questions

If $l o g_{2} (4^{x + 1} + 4) l o g_{2} (4^{x} + 1) = l o g_{2} 8,$ then x equals

log 2 (4^{x + 1} + 4) log 2 (4^{x} + 1) = log \frac{1}{\sqrt{2}} \sqrt{\frac{1}{8}}

{log}_{1 / \sqrt{2}} (\frac{1}{\sqrt{8}}) = {log}_{2} (4^{x} + 1) \times {log}_{2} (4^{x + 1} + 4)

x

l o g_{1 / \sqrt{2}} \frac{1}{\sqrt{8}} = l o g_{2} (4^{x} + 1) \times l o g_{2} (4^{x + 1} + 4)

x

{log}_{2} 8 = x

x

Join BYJU'S Learning Program