What is the value of #xy+yz+zx# if #x=1+log_a(bc)# , #y=1+log_b(ca)# , #z=1+log_c(ab) # ?

2 Answers
Oct 22, 2017

# xy+yz+zx = ( log^3(abc))/(loga logb logc) #

Where the base of the logarithm is arbitrary, all that is important is that we use the same base throughout).

Explanation:

We have:

# x = 1 + log_a(bc) # ..... [A]
# y = 1 + log_b(ca) # ..... [B]
# z = 1 + log_c(ab) # ..... [C]

We can change the base of the logarithms (the base is arbitrary, all that is important is that we use the same base throughout):

# y = log_n(x) => x = n^y #
# :. log x = log n^y #
# :. log x = ylog n #
# :. y = log x / log n #

Let #A=loga, B=logb# and #C=logc#

From Eq [A] we have:

# log_a(bc) = x-1 => x-1 = log (bc) / log a #
# :. x = (log b + logc) / log a + 1 #
# \ \ \ \ \ \ \ = (B + C) / A + 1 #
# \ \ \ \ \ \ \ = (A+B+C)/A #

From Eq [B] we have:

# log_b(ca) = y-1 => y-1 = log (ac) / log b #
# :. y = (log a + logc) / log b + 1 #
# \ \ \ \ \ \ \ = (A + C) / B + 1 #
# \ \ \ \ \ \ \ = (A+B+C)/B #

From Eq [C] we have:

# log_c(ab) = z - 1 => z-1 = log (ab) / log c #
# :. z = (log a + log b ) / log c + 1 #
# \ \ \ \ \ \ \ = (A +B) / C + 1 #
# \ \ \ \ \ \ \ = (A +B+C) / C #

So then the product we seek is:

# P = xy+yz+zx #
# \ \ \ = (A+B+C)/A (A+B+C)/B + (A+B+C)/B (A +B+C) / C + (A +B+C) / C (A+B+C)/A #

# \ \ \ = (A+B+C)^2/(AB) + (A+B+C)^2/(BC) + (A +B+C)^2/(AC) #

# \ \ \ = (C(A+B+C)^2 + A(A+B+C)^2 + B(A +B+C)^2)/(ABC) #

# \ \ \ = ( (A+B+C)^2(A+B+C))/(ABC) #

# \ \ \ = ( (A+B+C)^3)/(ABC) #

# \ \ \ = ( (loga+logb+logc)^3)/(loga logb logc) #

# \ \ \ = ( (log(abc))^3)/(loga logb logc) #

# \ \ \ = ( log^3(abc))/(loga logb logc) #

Oct 22, 2017

#x=1+log_a(bc)=log_aa+log_a(bc)#

#=>x=log_a(abc)#
#=>1/x=log_(abc)a#

Similarly

#=>1/y=log_(abc)b#

And

#=>1/z=log_(abc)c#

So

#1/x+1/y+1/z=log_(abc)a+log_(abc)b+log_(abc)c#

#=>(xy+yz+zx)/(xyz)=log_(abc)(abc)=1#

Hence

#xy+yz+zx=xyz#

#=(1+log_a(bc))(1+log_b(ca))(1+log_c(ab))#