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# If x = log a (bc), y = log b (ca) and z = log c (ab) prove that

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If x = loga(bc), y = logb(ca) and z = logc(ab) prove  that

+ y +  xy-  2.

Sep 29, 2017

#1
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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~

Sep 29, 2017
#2
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Note....    logd e  *  loge f =   log 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

Sep 29, 2017
edited by CPhill  Sep 29, 2017
edited by CPhill  Sep 29, 2017
edited by CPhill  Sep 29, 2017
edited by CPhill  Oct 4, 2017