In Solow model, on the balanced growth path, why aren't wages zero?

Question

According to this primer on the Solow Growth model, wages are equal to $w = \frac{\partial Y}{\partial L} = A \left(f(k) - f^{\prime}(k)k\right)$.

In the balanced growth path we know that $\dot{k} = 0 = sf(k^{\star}) - (n+g+\delta)k^{\star}$ where $k^{\star}$ is the value of $k$ on the balanced growth path.

From this it follows that

$$ f(k^{\star}) = \frac{n+g+\delta}{s}k^{\star} $$ $$ \implies f^{\prime}(k^{\star}) = \frac{n+g+\delta}{s} $$ $$ \implies f(k^{\star}) - f^{\prime}(k^{\star})k^{\star} = 0 $$

And so wages are zero.

But the primer also says that the growth rate of wages is $\frac{\dot{w}}{w} = g$ on the balanced growth path which doesn't make if we are dividing be $w=0$. So there must be a flaw in the above logic, but I don't see it.

Answer 1

The condition $$ f(k^\ast) = \frac{n + g + \delta}{s} k^\ast. $$ is an equilibrium condition which only holds at the value of $k^\ast$. As such, you are not allowed to take derivatives of both sides and equate them.

Taking derivatives would only be allowed if this equation holds for all $k$ in a neighbourhood of $k^\ast$ (which it doesn't).