Question

Based on this information answer the questions given
If $(a>1,x>1)$ or $(0<a<1,0<x<1)$.
then ${\log }_{a}x>0$ i.e., ${\log }_{a}x$ is positive.
If $(0<a<1,x>1)$ or $(a>1,0<x<1)$, then ${\log }_{a}x<0$
If $a>b$ then ${\log }_{b}a>1$ and ${\log }_{a}b<1.$

Determine the sign of $\frac{{\log }_a\left(3.162\right)}{{\log }_a\left(2/3\right)}.$

Answer 1

Given:

$A=\frac{{\log }_a\left(3.162\right)}{{\log }_a\left(2/3\right)}$

Let's consider $2$ cases:

Case-1: if $a>1$, then

${\log }_{a}3.162>0$; hence positive.

and ${\log }_a\left(2/3\right)={\log }_{a}0.67<0$; hence negative

So,

$A=\frac{{\log }_a\left(3.162\right)}{{\log }_a\left(2/3\right)}$ is a negative quantity.

Case-2: if $0<a<1$, then

${\log }_{a}3.162<0$; hence negative.

and ${\log }_a\left(2/3\right)={\log }_{a}0.67>0$; hence positive

So,

$A=\frac{{\log }_a\left(3.162\right)}{{\log }_a\left(2/3\right)}$ is again a negative quantity.

Hence sign of Case-1: if $a>1$, then

${\log }_{a}3.162>0$; hence positive.

and ${\log }_a\left(2/3\right)={\log }_{a}0.67<0$; hence negative

So,

$A=\frac{{\log }_a\left(3.162\right)}{{\log }_a\left(2/3\right)}$ is a negative quantity.

Hence sign of $A$ is negative.