CMAT 2016 Solved Question Paper

(d) ${\log }_2\left(9-2^{\mathit{x}}\right)={10}^{\log \left(3-\mathit{x}\right)}$
${\log }_2\left(9-2^{\mathit{x}}\right)$ = (3 - x) $\left[\begin{array}{c}\because \mathit{b}{\log }_{\mathit{b}}\left(\mathit{x}\right)=\mathit{a}\\ \Rightarrow {\mathit{b}}^{\mathit{a}}\end{array}\right]$
⇒ 9 - $2^{\mathit{x}}$ = $2^{\left(3-\mathit{x}\right)}$
⇒ 9 - $2^{\mathit{x}}$ = $\frac{2^3}{2^{\mathit{x}}}$ [ let $2^{\mathit{x}}$ = z]
⇒ 9- z = $\frac{8}{\mathit{z}}$
⇒ ${\mathit{z}}^2$ -9z + 8 = 0
⇒ (z - 1)(z - 8) = 0
⇒ z = 1 , 8
Now , $2^{\mathit{x}}$ = z
⇒ $2^{\mathit{x}}$ = 1
⇒ $2^{\mathit{x}}$ = $2^0$
∴ x = 0
Also, $2^{\mathit{x}}$ = z
⇒ $2^{\mathit{x}}$ = 8
⇒ $2^{\mathit{x}}$ = $2^3$
⇒ x = 3
So. the values of x can be 0 or 3.