Bank Exam Government Jobs Exams Numerical Ability Log and Antilog

If 2^x=3^y=6^(-z)
then what is the value of 1/x+1/y+1/z?

Read Solution (Total 1)

Let,2^x=3^y=6^-z=k
Takilog,log
X=log2(k)
y=log3(k)
-z=log6(k)
1/x+1/y+1/z=logk(2)+logk(3)-logk(6)
=logk(2*3)-logk(6)
=logk(6)-logk(6)
=0
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