Find the exact solution, in its simplest form, to the equation ln(4y + 7) = 3 + ln(2 – y) (Core Maths 3 Style Question)

First we subtract ln(2-y) from both sides to get "ln(4y+7) - ln(2-y) = 3". By applying the laws of logs from C2, we have "ln[(4y+7)/(2-y)] = 3". Therefore, if we put them as exponents of the base e, we have "e^{ln[(4y+7)/(2-y)]} = e³". By applying "lna = x means that e^x = a" (from C2) , we have that "e^{ln[(4y+7)/(2-y)]} = (4y+7)/(2-y)" so we get "(4y+7)/(2-y) = e³". Next, we multiply both sides by (2-y) to get "4y+7 = e³ (2-y)". If we expand the brackets on the right side, we have "4y+7 = 2e³-ye³". By collecting all the like terms on each side - y terms on left side and the remaining terms on the right, we get "4y+ye³ = 2e³-7". Now, we factorise the left so that we have "y(4+e³) = 2e³-7". Then, we can see that we can isolate y by dividing both sides by (4+e³) and so we have "y = (2e³-7)/(4+e³)". We cannot simplify this answer further so we have reached our final answer : y = (2e³-7)/(4+e³)