Question

The solution of the inequality ${\log }_x(2x-\frac{3}{4})>2$

• A. $x\in (\frac{3}{8},1)\cup (1,\frac{3}{2})$
• B. $x\in (1,\frac{3}{2})$
• C. $x\in (\frac{3}{8},\frac{1}{2})$
• D. $x\in (\frac{3}{8},\frac{1}{2})\cup (1,\frac{3}{2})$

Answer 1

The correct option is D $x\in (\frac{3}{8},\frac{1}{2})\cup (1,\frac{3}{2})$

${\log }_x(2x-\frac{3}{4})>2$
$\log$ function is defined when
$\left(i\right)x>0,x\ne 1$
$\left(ii\right)2x-\frac{3}{4}>0\Rightarrow x>\frac{3}{8}$
From $\left(i\right)$ and $\left(ii\right)$, we get
$x\in (\frac{3}{8},1)\cup (1,\infty )\dots \left(1\right)$

$\text{Case 1:}x\in (\frac{3}{8},1)\dots \left(2\right)$
${\log }_x(2x-\frac{3}{4})>2$
$\Rightarrow 2x-\frac{3}{4}<x^2$
$\Rightarrow 8x-3<4x^2$
$\Rightarrow 4x^2-8x+3>0$
$\Rightarrow (2x-3)(2x-1)>0$
$\Rightarrow x\in (-\infty ,\frac{1}{2})\cup (\frac{3}{2},\infty )\dots \left(3\right)$
Now, from $\left(2\right)\cap \left(3\right)$
$\Rightarrow x\in (\frac{3}{8},\frac{1}{2})\dots \left(4\right)$

$\text{Case 2:}x\in (1,\infty )\dots \left(5\right)$
${\log }_x(2x-\frac{3}{4})>2$
$\Rightarrow 2x-\frac{3}{4}>x^2$
$\Rightarrow 8x-3>4x^2$
$\Rightarrow 4x^2-8x+3<0$
$\Rightarrow (2x-3)(2x-1)<0$
$\Rightarrow x\in (\frac{1}{2},\frac{3}{2})\dots \left(6\right)$
Now, from $\left(5\right)\cap \left(6\right)$
$\Rightarrow x\in (1,\frac{3}{2})\dots \left(7\right)$

Now, $\left(4\right)\cup \left(7\right)$, gives
$\Rightarrow x\in (\frac{3}{8},\frac{1}{2})\cup (1,\frac{3}{2})$