+239 If x = loga(bc), y = logb(ca) and z = logc(ab) prove that
x + y + z = xyz - 2.
3 important logarithmic identities:
1.loga(b)=1/(logb(a))
2. loga(bc)=loga(b)+loga(c)
3. loga(b)*logb(c)=loga(c)
x+y+z= loga(bc)+logb(ca)+logc(ab)=(loga(b)+loga(c))+(logb(c)+logb(a))+(logc(a)+logc(b)).
d=loga(b)
e=loga(c)
f=logb(c)
x+y+z=(loga(b)+loga(c))+(logb(c)+logb(a))+(logc(a)+logc(b))=d+e+f+1/d+1/e+1/f
x*y*z-2=(loga(b)+loga(c))*(logb(c)+logb(a))*(logc(a)+logc(b))=(d+e)*(f+1/d)*(1/e*1/f)=(d*f)/e+d+1/e+1/f+f+e+1/d+e/(d*f)-2=d+e+f+1/d+1/e+1/f+((d*f)/e+e/(d*f)-2).
d*f=loga(b)*logb(c)=loga(c)=e therefore e/(f*d)=(d*f)/e=1 meaning:
x*y*z-2=d+e+f+1/d+1/e+1/f+((d*f)/e+e/(d*f)-2)=(d+e+f+1/d+1/e+1/f)+((d*f)/e+e/(d*f)-2)=(x+y+z)+(1+1-2)=(x+y+z)+0=x+y+z
therefore x+y+z=x*y*z-2
~blarney master~
+130081 Note.... logd e * loge f = logd f
Also ..... log e / log d + log f / log d = logd (ef)
Also........ [ log a / log c ] [ log c / log a ] = 1
Also ...........( logb c * logc b ) = 1
And..... logd e + logd f = logd (ef)
So......
xyz = [ logb / loga + log c / log a ] [ log c / log b + log a / log b ] [ log a / log c + log b / log c ] =
[ ( logb / loga)( log c / log b) + (logb / loga)( log a / log b) + ( log c / log a) ( log c / log b) +
(log c / log a) (log c / log a)] [ log a / log c + log b / log c ]
[ loga c + 1 + log c / loga * log c / log b + logb c ] [ logc a + logc b ] =
[ loga c + 1 + loga c * logb c + logb c ] [ logc a + logc b ] =
[ ( loga c) ( logc a) + (1)( logc a) + ( logc a * loga c) * logb c + (logb c)(logc a ) + ( loga c)( logc b) + (1)( logc b) + loga c *( logb c * logc b ) + ( logb c)(logc b ) ]
[ 1 + logc a + 1* logb c + logb a + loga b + logc b + loga c * 1 + 1 ] =
[ 2 + loga b + loga c + logb c + logb a + logc a + logc b ] =
2 + loga (bc) + logb(ca) + logc (ab)
2 + x + y + z ( subtract 2 from both sides)
xyz - 2 = x + y + z
