Number of solutions of equation ${\log }_2\left(9-2^x\right)={10}^{{\log }_{10}(3-x)}$ is :

We have,  

${\log }_2\left(9-2^x\right)={10}^{\log 10(3-x)}$

$\Rightarrow {\log }_2\left(9-2^x\right)=(3-x) \left[\because a^{{\log }_{a}N}=N\right]$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(9-2^x\right)=2^{(3-x)}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(9-2^x\right)=2^3\times 2^{-x}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9\cdot 2^x-{\left(2^x\right)}^2=8\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\left(2^x\right)}^2-9\cdot 2^x+8=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(2^x-8\right)\left(2^x-1\right)=0\end{array}$

$\Rightarrow 2^x=2^3$ or   $2^x=2$ $\Rightarrow x=3$or $x=0$

 Also, the given equation exists for

$9-2^x>0$ and $3-x>0$  

1.e. for  $2^x<9$ and $x<3$ 

1.e. for $x<{\log }_{2}9$ and $x<3$ 

Hence $, x = 0$  is the only solution of the given equation.