Let $f$ be a differentiable real-valued function defined on the positive real numbers. The tangent lines to the graph of $f$ always meet the $y$-axis 1 unit lower than where they meet the function. If $f(1)=0$, what is $f(2)$?
1. Let's find the tangent line to a point \((x_0, f(x_0))\). The slope of the tangent line at this point is given by \(f'(x_0)\). Using the point-slope form of the equation of a line, we have:
   \[
   y - f(x_0) = f'(x_0)(x - x_0)
   \]
   Converting this to slope-intercept form, we get:
   \[
   y = f'(x_0)x + f(x_0) - x_0 f'(x_0)
   \]
   Therefore, the y-intercept of this tangent line is:
   \[
   f(x_0) - x_0 f'(x_0)
   \]

2. We are given that the y-intercept of the tangent line is always 1 unit lower than where it meets the function. This means:
   \[
   f(x_0) - x_0 f'(x_0) = f(x_0) - 1
   \]
   Simplifying this equation, we find:
   \[
   -x_0 f'(x_0) = -1 \implies f'(x_0) = \frac{1}{x_0}
   \]

3. To find \(f(x)\), we integrate \(f'(x)\):
   \[
   f'(x) = \frac{1}{x} \implies f(x) = \int \frac{1}{x} \, dx = \ln(x) + C
   \]
   where \(C\) is the constant of integration.

4. We are given that \(f(1) = 0\). Substituting \(x = 1\) into the equation \(f(x) = \ln(x) + C\), we get:
   \[
   f(1) = \ln(1) + C = 0 \implies 0 + C = 0 \implies C = 0
   \]
   Therefore, the function \(f(x)\) is:
   \[
   f(x) = \ln(x)
   \]

5. Finally, we need to find \(f(2)\):
   \[
   f(2) = \ln(2)
   \]

The final answer is \(\boxed{\ln(2)}\)