Question

If $a>b$ and ${\log }_{b}a>1$, ${\log }_{a}b<1$, then the value of $\frac{{\log }_a\left(3.162\right)}{{\log }_a\left(\frac{2}{3}\right)}$ is

• A. $0$
• B. Negative
• C. Positive
• D. Cannot be determined

Answer 1

Answer: (B)

Negative

$\frac{{\log }_{a}3.162}{{\log }_a\left(\frac{2}{3}\right)}=\frac{-{\log }_{a}3.162}{-{\log }_a\left(\frac{2}{3}\right)}$
$=\frac{-{\log }_{a}3.162}{{\log }_a{\left(\frac{2}{3}\right)}^{-1}}$
$=\frac{-{\log }_{a}3.162}{{\log }_a\left(\frac{3}{2}\right)}$
$=-{\log }_{3/2}\left(3.162\right)[\because {\log }_{a}b=\frac{{\log }_{x}b}{{\log }_{x}a}]$
Now$,\frac{3}{2}=1.5<3.162$
$\therefore a=3.162$
$b=1.5$
${\log }_{b}a>1$
$\therefore {\log }_{3/2}\left(3.162\right)>1$
$\Rightarrow -{\log }_{3/2}\left(3.162\right)<-1$
Hence, the value is negative.