Asked by Denise
Express y as a function of x. C is a positive number.
3 ln y= 1/2ln (2x+1) - 1/3ln (x+4) + ln C
ln y = (1/6)ln (2x+1) -(1/9)ln (x+4) + (1/3) ln C
y = e^ln y = e^[(1/6)ln (2x+1) -(1/9)ln (x+4) + (1/3) ln C]
= C^(1/3)* (2x+1)^(1/6) /(x+4)^(1/9)
3 ln y= 1/2ln (2x+1) - 1/3ln (x+4) + ln C
ln y = (1/6)ln (2x+1) -(1/9)ln (x+4) + (1/3) ln C
y = e^ln y = e^[(1/6)ln (2x+1) -(1/9)ln (x+4) + (1/3) ln C]
= C^(1/3)* (2x+1)^(1/6) /(x+4)^(1/9)