Find the exact solution to ln(2y+5) = 2 + ln(4-y)

1. To solve the equation ln(2y + 5) = 2 + ln(4 – y), we begin by isolating the logarithmic terms. We can utilise the properties of logarithms to combine the terms on the right side of the equation. By rewriting the equation, we have ln(2y + 5) – ln(4 – y) = 2. This can be further simplified using the quotient rule of logarithms, which states that ln(a) – ln(b) = ln(a/b). Thus, we can express the equation as ln((2y + 5) / (4 – y)) = 2.

2. Next, we eliminate the natural logarithm by exponentiating both sides of the equation. This leads us to the equation (2y + 5) / (4 – y) = e^2. To solve for y, we will cross-multiply, resulting in 2y + 5 = e^2(4 – y). Expanding this gives us 2y + 5 = 4e^2 – e^2y. We can then rearrange the terms to isolate y on one side of the equation, yielding, 2y + e^2y = 4e^2 – 5.

3. Finally, we factor out y from the left side, resulting in y(2 + e^2) = 4e^2 – 5. To find the value of y, we divide both sides by (2 + e^2), leading us to the solution y = (4e^2 – 5) / (2 + e^2). This process exemplifies the type of problem-solving skills that are essential for success in October A Level Maths Revision, where understanding logarithmic functions and their properties is crucial for tackling complex equations effectively.