Question

If ,$\frac{{\log }_{a}N}{{\log }_{c}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{b}N-{\log }_{c}N},$ where $N>0$ and $N\ne 1,a,b,c>0$ and not equal to 1, then

• A. $b^2=ac$
• B. $b^2=2ac$
• C. $b^2=4ac$
• D. $b^2=-ac$

Answer 1

The correct option is A $b^2=ac$

$\frac{{\log }_{a}N}{{\log }_{c}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{b}N-{\log }_{c}N}$

$\Rightarrow \frac{\frac{\log N}{\log a}}{\frac{\log N}{\log c}}=\frac{\frac{\log N}{\log a}-\frac{\log N}{\log b}}{\frac{\log N}{\log b}-\frac{\log N}{\log c}}$

$\Rightarrow \frac{\log c}{\log a}=\frac{(\log b-\log a)\left(\log b\log c\right)}{(\log c-\log b)\left(\log a\log b\right)}$

$\Rightarrow \log c-\log b=\log b-\log a\Rightarrow \log c+\log a=2\log b\Rightarrow b^2=ac$