If ${log}_{2} (30) + (x - 4) - 2 {log}_{2} (1 - 2^{x - 4}) = - {log}_{2} (0.5 - 2^{x - 5})$ , then the value of $x$ is

Open in App

Solution

${log}_{2} (30) + (x - 4) - 2 {log}_{2} (1 - 2^{x - 4}) = - {log}_{2} (0.5 - 2^{x - 5})$
$\Rightarrow {log}_{2} (2 \times 15) + (x - 4) - 2 {log}_{2} (1 - 2^{x - 4}) = - {log}_{2} [0.5 (1 - 2^{x - 4})]$
$\Rightarrow 1 + {log}_{2} (15) + (x - 4) - 2 {log}_{2} (1 - 2^{x - 4}) = - {log}_{2} (0.5) - {log}_{2} (1 - 2^{x - 4})$
$\Rightarrow 1 + {log}_{2} (15) + (x - 4) = 1 + {log}_{2} (1 - 2^{x - 4})$
$\Rightarrow {log}_{2} (1 - 2^{x - 4}) - {log}_{2} (15) = x - 4$
$\Rightarrow {log}_{2} (\frac{1 - 2^{x - 4}}{15}) = x - 4$
$\Rightarrow 1 - 2^{x - 4} = 15 \cdot 2^{x - 4}$
$\Rightarrow 1 = 16 \cdot 2^{x - 4}$
$\Rightarrow 1 = 2^{4} \cdot 2^{x - 4}$
$\Rightarrow 2^{0} = 2^{x}$
$\Rightarrow x = 0$

Suggest Corrections

Similar questions

∣ ∣ ∣ ∣ \begin{matrix} x + 4 & 2 x & 2 x 2 x & x + 4 & 2 x 2 x & 2 x & x + 4 \end{matrix} ∣ ∣ ∣ ∣, x \in R - {\frac{- 4}{5}, 4}

2 (1 + 2 x) - 5 (4 - 2 x) = 14

x

\frac{3 x - 1}{4} - \frac{2 x + 5}{3} = \frac{5}{2} - 2 x,

f (x) = \sqrt{x + 2 \sqrt{2 x - 4}} + \sqrt{x - 2 \sqrt{2 x - 4}}

10 f^{'} (102^{+})

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 6 - {cos}^{2} x & {cos}^{2} x & 4 sin 2 x {sin}^{2} x & 5 + {cos}^{2} x & 4 sin 2 x {sin}^{2} x & {cos}^{2} x & 5 + 4 sin 2 x \end{matrix} ∣ ∣ ∣ ∣ ∣,

x \in R .

M

m

f

M - m

Join BYJU'S Learning Program