Question

If log a base ab = x. Then prove that log ab base b =1/(1-x)

Answer 1

Answer:

THE ANSWER IS HERE,

= > \: log_{a}(ab) = x=>loga(ab)=x

= > \: log_{a}(a) + log_{a}(b) = x=>loga(a)+loga(b)=x

= > \: 1 + log_{a}(b) = x=>1+loga(b)=x

= > \: log_{a}(b) = x - 1=>loga(b)=x−1

Given that,

= > \: log_{b}(ab)=>logb(ab)

= > \: log_{b}(a) + log_{b}(b)=>logb(a)+logb(b)

= > \: \frac{1}{ log_{a}(b) } + 1=>loga(b)1+1

= > \: \frac{1}{x - 1} + 1=>x−11+1

= > \: \frac{1 + x - 1}{x - 1}=>x−11+x−1

= > \: \frac{x}{x - 1}=>x−1x