The sum of values of $x$ satisfying ${log}_{1 / 2} (x - 1) + {log}_{1 / 2} (x + 1) - {log}_{1 / \sqrt{2}} (7 - x) = 1$ is

Open in App

Solution

${log}_{1 / 2} (x - 1) + {log}_{1 / 2} (x + 1) - {log}_{1 / \sqrt{2}} (7 - x) = 1$
$\Rightarrow - {log}_{2} (x - 1) - {log}_{2} (x + 1) + {log}_{\sqrt{2}} (7 - x) = 1$
For the $log$ to be defined,
$x - 1 > 0 \Rightarrow x > 1 x + 1 > 0 \Rightarrow x > - 1 7 - x > 0 \Rightarrow x < 7 ∴ x \in (1, 7)$

Now, $- {log}_{2} (x - 1) - {log}_{2} (x + 1) + {log}_{\sqrt{2}} (7 - x) = 1$
$\Rightarrow - {log}_{2} (x - 1) - {log}_{2} (x + 1) + 2 {log}_{2} (7 - x) = 1 \Rightarrow {log}_{2} \frac{(7 - x)^{2}}{(x + 1) (x - 1)} = 1 \Rightarrow \frac{(x - 7)^{2}}{(x + 1) (x - 1)} = 2 \Rightarrow x^{2} - 14 x + 49 = 2 x^{2} - 2 \Rightarrow x^{2} + 14 x - 51 = 0 \Rightarrow (x + 17) (x - 3) = 0 \Rightarrow x = - 17, 3$
But $x \neq - 17$
$∴ x = 3$

Suggest Corrections

Similar questions

x

{log}_{1 / 3} (2 {(\frac{1}{2})}^{x} - 1) = {log}_{1 / 3} ({(\frac{1}{4})}^{x} - 4)

x

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

S

x

l o g_{1 / 2} (l o g_{2} (x + \sqrt{5}) + l o g_{2} (x - \sqrt{5}) + l o g_{1 / 2} (x - 2)) - 1

S

{log}_{1 / 2} (4 - x) \geq {log}_{1 / 2} 2 - {log}_{1 / 2} (x - 1)

x \in

{log}_{\frac{1}{2}} (x - 1) + {log}_{\frac{1}{2}} (x + 1) - {log}_{\frac{1}{\sqrt{2}}} (7 - x) = 1

Join BYJU'S Learning Program