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)

${\log }_2\left(9-2^x\right)=10^{{\log }_{10}(3-x)}$
$\Rightarrow {\log }_2\left(9-2^x\right)=(3-x)$
$\left[\because a^{{\log }_{a}b}=b\right]$
$\Rightarrow 2^{3-x}=9-2^x$
$\Rightarrow \frac{2^3}{2^x}=9-2^x$
$\Rightarrow 8=2^x\times \left(9-2^x\right)$
$\Rightarrow 2^{2x}-2^x\times 9+8=0$
Let $2^x=y$, then;
$y^2-9y+8=0$
$\Rightarrow (y-8)(y-1)=0$
$\Rightarrow y=8$ 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.