Question

If logabc  =  x, logbca  =  y and logcab  =  z, then find the value of

1/(x+1) + 1/(y+1) + 1/(z+1)

​

Answer 1

Solution :

x + 1 = logabc + logaa = logaabc

y + 1 = logbca + logbc = logbabc

z + 1 = logcab + logcc = logcabc

1/(x+1) = 1 / logaabc = logabca

1/(y+1) = 1/logbabc = logabcb

1/(z+1) = 1/logcabc = logabcc

1/(x+1) + 1/(y+1) + 1/(z+1) = logabca + logabcb + logabcc

1/(x+1) + 1/(y+1) + 1/(z+1) = logabcabc

1/(x+1) + 1/(y+1) + 1/(z+1) = 1