If ${log}_{2} (2^{x - 1} + 6) + {log}_{2} (4^{x - 1}) = 5$ , then $x =$

4

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

1

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

3

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

Open in App

Solution

The correct option is C $2$
${l o g}_{2} (2^{x - 1} + 6) + {l o g}_{2} (4^{x - 1}) = 5$

$⟹ {log}_{2} (2^{x - 1} + 6) (4^{x - 1}) = 5 {log}_{2} 2$
$⟹ {log}_{2} (2^{x - 1} + 6) (4^{x - 1}) = {log}_{2} 2^{5}$

$⟹ (2^{x - 1} + 6) (4^{x - 1}) = 32$

Let $2^{x - 1} = k$

$⟹ (k + 6) (k^{2}) = 32$
$⟹ k^{3} + 6 k^{2} - 32 = 0$
$⟹ k = 2, - 4 ⟹ 2^{x - 1} = 2$ but $2^{x - 1} \neq - 4$

$⟹ x = 2$

Suggest Corrections

Similar questions

{log}_{2} (4^{x} + 1) = x + {log}_{2} (2^{x + 3} - 6)

{log}_{2} (4^{x} + 4) = {log}_{2} 2^{x} + {log}_{2} (2^{x + 1} - 3)

l o g_{2} (4^{x} + 4) - l o g_{2} (2 x + 1 - 3) = 0.

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

x

\sqrt{{log}_{2} (\frac{2 x - 3}{x - 1})} < 1,

x \in

Join BYJU'S Learning Program