The equation \[{\log _2}(3 - x) + {\log _2}(1 - x) = 3\]has
A.One root
B.Two roots
C.Infinite roots
D.No root

Hint: Here we will use the properties of log we have to remember this
Property of ‘Log’
I-Property: -
\[\log a + \log b = \log (ab)\]
II-Property: -
\[log\,ab = x\]
\[b = {a^x}\]
\[\log ab\] is called \[\log b\] with base


Complete step by step solution:
Given
Equation: -
\[{\log _2}(3 - x) + {\log _2}(1 - x) = 3\]
Using property of ‘log’ i.e.\[\log a + \log b = \log (ab)\], we get:
\[ \Rightarrow {\log _2}(3 - x)(1 - x) = 3\]
On multiplying the brackets, we get:
\[ \Rightarrow {\log _2}(3 - 3x - x + {x^2}) = 3\]
\[ \Rightarrow {\log _2}(3 - 4x + {x^2}) = 3\]
Now using property of log
\[ \Rightarrow \log ab = x\]
\[ \Rightarrow b = {(a)^x}\] _____equation (1)
Here \[b = 3 - 4x + {x^2}\]
\[a = 2\]
\[x = 3\]
Put these values in equation (1)
\[ \Rightarrow 3 - 4x + {x^2} = {(2)^3}\]
\[3 - 4x + {x^2} = 8\]
\[{x^2} - 4x + 3 - 8 = 0\] {On transposing 8 on L.H.S}
\[{x^2} - 4x - 5 = 0\] __________equation (2)
Now factorising this equation, we get:
Write \[4x\,as\,( - 5x + x)\]in equation (2)
\[{x^2} - 5x + x - 5 = 0\]
\[x(x - 50 + 1(x - 5) = 0\] {taking x common from first bracket&\[ + 1\] common from second bracket}
\[(x - 5)(x + 1) = 0\] {factors of equations}
\[x = 5\,or\,x = - 1\]
\[x = 1\]
Hence required roots are one root because ‘log’ cannot be negative.
So, option A is the right answer


Note: Value of \[x = - 1\,\& \,5\]
Our responsibility is to check that these two values \[ - 1\,\& 5\] are in its domain or not.
So, we see it when we put \[x = 5\]in a given \[e{q^n}\], log becomes negative.
“log is never negative “
So, the final root is \[1\].