Logarithm Question

If $\log_a c = \cfrac{\log_a N - \log_b N}{\log_b N - \log_c N} $ where N 0 , N $\ne$ 1 .

a,b,c greater than zero... and a $\ne $ 1 , b $\ne$ 1 , c $\ne$ 1 , a $\ne$ c ... then prove that $b^2 = ac$

Question

Logarithm Question

If $\log_a c = \cfrac{\log_a N - \log_b N}{\log_b N - \log_c N} $ where N > 0 , N $\ne$ 1 . 

a,b,c greater than zero... and a $\ne $ 1 , b $\ne$ 1 , c $\ne$ 1 , a $\ne$ c ... then prove that $b^2 = ac$

ganeshie8 · Answer

componendo and dividendo comes to my mind

hartnn · Answer

log N gets cancelled from numerator and denominator if you use 
log_x y = log y/ log x

mathslover · Answer

Oh, fine.. changing the base worked...

mathslover · Answer

$\log_a c $ cancelled out... and it became easy.. then.

mathslover · Answer

thanks @ganeshie8 and @hartnn