Question

The true solution set of inequality $lo{g}_{(2x-3)}(3x-4)>0$ is equal to:

• A. $(\frac{4}{3},\frac{5}{3})\cup (2,\infty )$
• B. $(\frac{3}{2},\frac{5}{3})\cup (2,\infty )$
• C. $(\frac{4}{3},\frac{3}{2})\cup (2,\infty )$
• D. $(\frac{2}{3},\frac{4}{3})\cup (2,\infty )$

Answer 1

The correct option is B $(\frac{3}{2},\frac{5}{3})\cup (2,\infty )$

Given ${\log }_{(2x-3)}(3x-4)>0$

$\frac{\log (3x-4)}{\log (2x-3)}>0$

Either both numerator and denominator are positive and both arre negative.

$0<2x-3<1$

$\frac{3}{2}<x<2$

and $0<3x-4<1$

$\frac{4}{3}<x<\frac{5}{3}$

Taking intersection

for, $xϵ(\frac{3}{2},\frac{5}{3})$

$\log (3x-4)$ and $\log (2x-3)$ both are negative,

Hence, $\frac{\log (3x-4)}{\log (2x-3)}>0$

for, $xϵ(2,\infty )$

Both $\log (2x-3)$ and $\log (3x-4)$ both are positive,

Hence, $\frac{\log (3x-4)}{\log (2x-3)}>0$

Hence, $xϵ(\frac{3}{2},\frac{5}{3})\bigcup (2,\infty )$