If Logx-6 base3=2y and Logx-7 base2=3y, show that x^2-13x+42=72y

Answer 1

You have a typo.

the right side of your equation should say 72^y

log₃(x-7) = 2y --> 3^(2y) = x-7
log₂(x-6) = 3y --> 2^(3y) = x-6

then
LS = x^2 - 13x + 42
= (x-6)(x-7)
= ( 2^(3y) ) ( 3^(2y) )
= (8^y)(9^y)
= 72^y
= RS