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

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.

## 1 Answer

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).