Question

Let $a>0,c>0,b=\sqrt{ac},a,c$ and $ac\ne 1,N>0$
Prove that $\frac{1}{2}\frac{{\log }_{a}N}{{\log }_{c}N}=\frac{{\log }_{a}N-{\log }_{b}N}{{\log }_{b}N-{\log }_{c}N}$

Answer 1

Using Base change theorem, we get
RHS $=\frac{\frac{1}{{\log }_{N}a}-\frac{1}{{\log }_{N}c}}{\frac{1}{{\log }_{N}b}-\frac{1}{{\log }_{N}c}}$
$=\frac{\frac{{\log }_{N}b-{\log }_{N}a}{{\log }_{N}b{\log }_{N}a}}{\frac{{\log }_{N}c-{\log }_{N}b}{{\log }_{N}c{\log }_{N}b}}$
Using Quotient law
$=\frac{{\log }_N\left(\frac{b}{a}\right)}{{\log }_{N}a}\frac{{\log }_{N}c}{{\log }_N\left(\frac{c}{b}\right)}$
We have $b=\sqrt{ac}\Rightarrow \sqrt{b}\sqrt{b}=\sqrt{ac}$..........$\left(1\right)$
$\Rightarrow \sqrt{\frac{b}{a}}=\sqrt{\frac{c}{b}}$
Substituting the above in eqn$\left(1\right)$ we get
$\frac{{\log }_N\left(\sqrt{\frac{c}{b}}\right)}{{\log }_{N}a}\frac{{\log }_{N}c}{{\log }_N\left(\frac{c}{b}\right)}=\frac{1}{2}\frac{{\log }_{N}c}{{\log }_{N}a}$
$=\frac{1}{2}{\log }_{a}c=$ L.H.S