If `2^(x+y) = 6^(y) and 3^(x-1) = 2^(y+1)`, then the value of `(log 3 - log 2)//(x-y)` is

Options

A. 1
B. `log_(2)3-log_(3)2`
C. ` log(3//2)`
D. none of these

Answer

C

Explanation

Taking log, we have
`(x+y) log 2 = y (log 2+ log 3)`
`:. x log 2 = y log 3`
` or x/(log 3) = y/(log 2) = (x-y)/(log 3 - log 2) = lambda`(say) …(i)
Also ` (x-1) log 3 = (y+1) log 2`
` or x log 3 - y log 2 = log 3 + log 2`
Using Eq. (i), we get
` lambda[(log 3)^(2)-(log 2)^(2)] = log 3+log 2`
` lambda = 1/(log 3-log 2)`
` :. 1/lambda = log 3 - log 2 = log. 3/2`