Question

$\log _{2}\left(9-2^{x}\right)=10^{\log (3-x)}$, solve for $x$

• A. 0
• B. 3
• C. both (a) and (b)
• D. 0 and 6

Answer 1

Answer: (A)

$\log _{2}\left(9-2^{x}\right)=10^{\log _{10}(3-x)}$
$\Rightarrow \log _{2}\left(9-2^{x}\right)=(3-x) $
$\left[\because a^{\log _{a} b}=b\right]$
$\Rightarrow 2^{3-x}=9-2^{x}$
$\Rightarrow \frac{2^{3}}{2^{x}}=9-2^{x}$
$\Rightarrow 8=2^{x} \times\left(9-2^{x}\right)$
$\Rightarrow 2^{2 x}-2^{x} \times 9+8=0$
Let $2^{x}=y$, then;
$y^{2}-9 y+8=0$
$\Rightarrow (y-8)(y-1)=0$
$\Rightarrow y=8$ or $y=1$
$\Rightarrow 2^{x}=2^{3}$ or $2^{x}=2^{0}$
$\Rightarrow x=3$ or $x=0$.
But $x=3$ does not satisfy the given equation, since $\log 0$ is not defined.