Question

a powerx=(a\b)whole powery=k power m then 1/x-1/y?

Answer 1

Given,
a^x = (a/b)^y = k^m

a^x = k^m
taking log both sides,
Loga^x = logk^m
xloga = mlogk
x = mlogk/loga --------(1)

Similarly, (a/b)^y = k^m
Taking log both sides,
log(a/b)^y = logk^m
ylog(a/b) = mlogk
y = mlogk/log(a/b) ---------(2)

Now, 1/x - 1/y
Putting equations (1) and (2) ,
loga/mlogk - log(a/b)/mlogk
= (loga - log(a/b))/mlogk
= (loga - loga + logb)/mlogk
= logb/mlogk
= 1/m logb/logk

Hence, 1/x - 1/y = 1/m $Log_k^b$