We have an equation \[{\log _2}^{\left( {9 - {2^x}} \right)} = {10^{{{\log }^{(3 - x)}}}}\]. Solve for \[x\].
a). 0
b). 3
c). Both (a) and (b)
d). 0 and 6

Formula used:  \[{a^{{{\log }_a}b}} = b\]

Complete step-by-step solution:
Given; logarithmic equation is \[{\log _2}^{\left( {9 - {2^x}} \right)} = {10^{{{\log }^{(3 - x)}}}}\]
To find the value of \[x\]
\[{\log _2}^{\left( {9 - {2^x}} \right)} = {10^{{{\log }^{(3 - x)}}}}\]
We know that \[{a^{{{\log }_a}b}} = b\]
\[{\log _2}^{\left( {9 - {2^x}} \right)} = 3 - x\]
Taking exponent on 2
\[{2^{{{\log }_2}^{\left( {9 - {2^x}} \right)}}} = {2^{\left( {3 - x} \right)}}\]
Using same property of logarithmic
\[9 - {2^x} = \dfrac{{{2^3}}}{{{2^x}}}\]
On further simplifying
\[{2^x}\left( {9 - {2^x}} \right) = {2^3}\]
On further solving
\[9 \times {2^x} - {2^{2x}} = 8\]
On rearranging
\[{2^{2x}} - 9 \times {2^x} + 8 = 0\]
Let \[{2^x} = t\] , then on putting the values,
\[{t^2} - 9t + 8 = 0\]
Solving by factorization method
On factoring
\[{t^2} - t - 8t + 8 = 0\]
On taking common terms
\[t\left( {t - 1} \right) - 8\left( {t - 1} \right) = 0\]
On taking again common
\[(t - 1)(t - 8) = 0\]
On equating both the terms to 0
\[t = 1\] ……………………(i)
\[t = 8\] ……………..……(ii)
On putting the values of t in equation (i)
\[{2^x} = 1\]
On writing expression like
\[{2^x} = {2^0}\]
On comparing both the terms
\[x = 0\]
On putting the values of t in equation (i)
\[{2^x} = 1\]
On writing expression like
\[{2^x} = {2^0}\]
On comparing both the terms
\[x = 0\]
On putting the values of t in equation (ii)
\[{2^x} = 8\]
On writing expression like
\[{2^x} = {2^3}\]
On comparing both the terms
\[x = 3\]
The values of $x$ are
\[ \Rightarrow x = 0\] and
\[ \Rightarrow x = 3\]
But \[x = 3\] not supports the equation because \[log{\text{ }}0\] not defined
The value of \[x\] that satisfies the equation is $x=0$.

Therefore, Option (a) is the correct answer.

Note: After finding the values of \[x\], we must check the obtained answer because some of the answers are not satisfying the parental equation. We have to observe the base of the logarithmic function carefully. If the base of the logarithmic function is not given then we use that as base $10$ by default.