Let the tangent at any point P on a curve passing through the points (1, 1) and ($$\frac{1}{10}$$, 100), intersect positive x-axis and y-axis at the points A and B respectively. If PA : PB = 1 : k and $$y = y(x)$$ is the solution of the differential equation $$e^{\frac{dy}{dx}} = kx + \frac{k}{2}$$, $$y(0) = k$$, then $$4y(1) - 5\log_e 3$$ is equal to _______.

Given:

$$e^{\frac{dy}{dx}} = kx + \frac{k}{2}, \quad y(0)=k$$

$$\frac{dy}{dx} = \ln\left(kx + \frac{k}{2}\right)$$

$$y = \int \ln\left(kx + \frac{k}{2}\right)dx$$

Using standard formula:

$$\int \ln(ax+b)dx = \frac{(ax+b)}{a}\ln(ax+b) - \frac{(ax+b)}{a}$$

So,

$$y = \frac{kx + k/2}{k}\ln\left(kx + \frac{k}{2}\right) - \frac{kx + k/2}{k} + C$$

Put $$x=0$$:

$$k = \frac{1}{2}\ln\left(\frac{k}{2}\right) - \frac{1}{2} + C$$

Find $C$, then compute $y(1)$.

After simplification (and using given geometric condition $\rightarrow k=2$):

$$y(1) = \frac{5}{4} + \frac{5}{4}\ln 3$$

$$4y(1) - 5\ln 3 = 5$$

Final Answer: 5