Question

Log (3y+2)-1=log(y-4)

Answer 1

y=6

Explanation:

`\log(3y + 2) - 1 = \log(y - 4)`

`\implies \log(3y + 2) - \log(10)= \log(y - 4)`

`\implies \displaystyle\log( (3y + 2)/(10)) = \log(y - 4)`

`\implies (3y + 2)/(10) = y - 4`

`\implies 3y + 2 = 10y - 40`

`\implies 7y = 42`

`\boxed{ y = 6}`