Prove that:
[log_base a_(x)] / [log_base ab_(x)] = 1 + log_base a_(b), for all a, b >1 and x >0.
Thanks
Hello, Mr_Green!
If we're allowed the Base-Change Formula, it's not bad . . .
We will change the denominator to base a.Prove that: .[log_a(x)] / [log_ab(x)] .= . 1 + log_a(b)
. . . . . . . . .log_a(x) . . . . . .log_a(x)
We have: . ----------- . = . -------------- . = . log_a(ab)
. . . . . . . . log_ab(x) . . . . . log_a(x)
. . . . . . . . . . . . . . . . . . . .-----------
. . . . . . . . . . . . . . . . . . . .log_a(ab)
And: . log_a(ab) . = . log_a(a) + log_a(b) . = . 1 + log_a(b)