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 \(b^2 = ac\)

Answer 1

componendo and dividendo comes to my mind

Answer 2

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

Answer 3

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

Answer 4

\(\log_a c \) cancelled out... and it became easy.. then.

Answer 5

thanks @ganeshie8 and @hartnn