If x=loga(bc), y=logb(ca), and Z=logc(ab). Prove that x+y+Z=xyZ-2

Answer 1

If we change bases to make all those logs base e, we have

x = ln(bc)/lna
y = ln(ac)/lnb
z = ln(ab)/lnc

x+y+z = (ln(bc)*lnb*lnc + ln(ac)*lna*lnc + ln(ab)*lna*lnb]/(lna*lnb*lnc)

xyz-2 = (ln(bc)*ln(ac)*ln(ab)-2*lna*lnb*lnc)/(lna*lnb*lnc)

OK. Just to make things more readable, let's dispense with all the "ln" stuff, recognize that there is a common denominator, so we just have to work with the numerators:

x+y+z = (b+c)bc+(a+c)ac+(a+b)ab
xyz-2 = (b+c)(a+c)(a+b)-2abc

If you expand both of those products and sums, you will see that they are indeed equal. Yes, I have done it, and it's not too hard. Just write things down carefully.