Question

log(4y-3)=log(2y+1)-log3​

Answer 1

Step-by-step explanation:

log(4y-3)=log(2y+1)-log3

log3=log(2y+1)-log(4y-3)

log3=log( (2y+1)/(4y-3) )

so

3=(2y+1)/(4y-3)

3×(4y-3)=(2y+1)

12y-9=2y+1

12y-2y=1+9

10y=10

so y=1