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
# 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
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) #
Similarly
And
So
Hence
