Question

logx^a= logy^b=log(yx)^ab show that a+b=1

Answer 1

\[\log(xy)^{ab}=ab \log(xy)=ab(\log x + \log y)=ab \log x+ab \log y\]\[=b \log x^a+a \log b^y = b \log y^b +a \log y^b =(a+b) \log y ^b\]
So we have shown that:\[\log (xy)^{ab}=(a+b)\log y^b \],but we also have:\[\log y^b = \log (xy)^{ab}\]so that gives us:\[\log y^b = (a+b) \log y^b\]and by dividing both sides by log y^b we obtain:\[a+b=1\]