Question

Find the set of real values of x for which ${\log }_2$ ($x^2$ - x - 6) + ${\log }_{0.5}$ (x - 3) < 2${\log }_{2}3$.

• A. x ∈ (3, ∞)
• B. x ∈ (3, 7)
• C. x ∈ (-∞, 7)
• D. x ∈ (0, 7)

Answer 1

The correct option is B

x ∈ (3, 7)

${\log }_2$ ($x^2$ - x - 6) + ${\log }_{0.5}$ (x - 3) < 2${\log }_{2}3$

For log to be defined

Condition 1: $x^2$ - x - 6 > 0

(x + 2) (x - 3) > 0

x ∈ ($-\infty$ , -2) U(3,$\infty$) - - - - - - (1)

Condition 2: x - 3 > 0

x > 3

x ∈ (3,$\infty$) - - - - - - (2)

Over all conditions for above logarithms to be defined is the common part of both the conditions

x ∈ (3, $\infty$) - - - - - - (3)

${\log }_2$ ($x^2$ - x - 6) + ${\log }_{0.5}$ (x - 3) < 2${\log }_{2}3$

Make same base for all logarithmic terms

${\log }_2$ (x + 2) + ${\log }_2$ (x - 3) - ${\log }_2$ (x - 3) < 2 ${\log }_2$ 3

${\log }_2$ (x+2) < ${\log }_2$ 9

Base of the log is greater than 1, then inequality is equivalent to

Then x + 2 < 9

x < 7

x ∈ ($-\infty$, 7) - - - - - - (4)

But from conditions of log we got, x ∈ (3,$\infty$)

Common port of equation 3 and 4 can be calculated based on number line

Common part is (3, 7)