Question

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

Answer 1

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 $1$.