Number of solutions of equation log_2 (9-2^x)=10^log_10(3-x), is ____…

We have,

$\log_2 (9-2^x)=10^{\log_{10}(3-x)}$

$⇒ \log_2 (9-2^x)=(3-x)$ $[∵a^{\log_a\, N}=N]$

$⇒(9-2^x)=2^{(3-x)}$

$⇒(9-2^x)=2^3×2^{-x}$

$⇒9.2^x-(2^x)2=8$

$⇒(2^x)^2-9.2^x+8=0$

$⇒(2^x-8) (2^x-1)=0$

$⇒2^x = 2^3$ or, $2^x = 2^0⇒ x=3$ or, $x=0$

Also, the given equation exists for

$9-2^x>0$ and $3-x>0$

i.e. for $2^x <9$ and $x < 3$

i.e. for $x <\log_2 9$ and $x < 3$.

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