Question

Find the real values of x satisfying ${\log }_{0.3}$ (10x + 3) < ${\log }_{0.3}$ (7x - 4).

• A. x ∈
• B.
• C. x ∈
• D.

Answer 1

The correct option is A

x ∈

${\log }_{0.3}$ (10x + 3) < ${\log }_{0.3}$ (7x - 4)

For log to be defined

\begin{array}{c|c} \hline
\text{10x + 3 > 0} & \text{7x - 4 > 0} \\\hline

\text{10x > -3} & \text{7x > 4} \\\hline

\text{x > $\frac{-3}{10}$} & \text{x > $\frac{4}{7}$} \\ \hline
\end{array}

On a number line

Common region is x ∈ $(\frac{4}{7},\infty )$

${\log }_{0.3}$ (10x + 3) < ${\log }_{0.3}$ (7x - 4)

Since 0.3 < 1 {base of the log is lies between 0 to 1}

Inequality is equivalent to
$10x+3>7x-4$
$3x>-7$
$x>\frac{-7}{4}$
x ∈ $(\frac{4}{7},\infty )$