Question

If $\frac{{\log }_{a}N}{{\log }_{c}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{b}N-{\log }_{c}N}$, where $N>0$ & $N\ne 1,a,b,c>0$ & $\ne 1$, then $b^2$ is equal to

• A. $ac$
• B. $\frac{a}{c}$
• C. $\frac{a^2}{c^2}$
• D. $a^2{c}^2$

Answer 1

The correct option is A $ac$

$\frac{{\log }_{a}N}{{\log }_{c}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{b}N-{\log }_{c}N}$
$\Rightarrow \frac{{\log }_{N}c}{{\log }_{N}a}=\frac{\frac{1}{{\log }_{N}a}-\frac{1}{{\log }_{N}b}}{\frac{1}{{\log }_{N}b}-\frac{1}{{\log }_{N}c}}[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
$\Rightarrow \frac{{\log }_{N}c}{{\log }_{N}a}=\frac{{\log }_{N}c\left({\log }_{N}b-{\log }_{N}a\right)}{{\log }_{N}a\left({\log }_{N}c-{\log }_{N}b\right)}$
$\Rightarrow {\log }_N\frac{b}{a}={\log }_N\frac{c}{b}$
$\Rightarrow \frac{b}{a}=\frac{c}{b}\Rightarrow b^2=ac$