Question

Find the real values x satisfying ${\log }_{10}$ x+ ${\log }_{10}$ (2-x) < 1 .

• A. x ∈ (0, 2)
• B. All real values
• C. x ∈ (0, 1)
• D. x < 2

Answer 1

The correct option is A

x ∈ (0, 2)

Given inequality ${\log }_{10}$ x+ ${\log }_{10}$​ (2 - x) < 1 ----------------(1)

For log to be defined x > 0

2 - x > 0

x - 2 < 0

x < 2

$\Rightarrow$ x ∈ (0,2) ------------------------(2)

Since ${\log }_{10}$ x+ ${\log }_{10}$​ (2 - x) < 1

${\log }_{10}x$​ (2 - x) < 1

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

x(2 - x) < ${10}^1$

2x - $x^2$ - 10 < 0

$x^2$ - 2x + 10 > 0

$(x-1{)}^2$ + 9 > 0 ---------------------------(3)

This is true for all values of x

From equation 2 & 3

x ∈ (0,2)