Proving a result In Jones (1999) "Growth: With or Without Scale Effects"
We know that $g_A = \dot{A}/A$ and thus $g_A = \delta L_A A^{\phi-1}$.

Take logs of both sides of $g_A = \delta L_A A^{\phi-1}$ and differentiating with respect to time gives us

$$ \frac{\dot{g_A}}{g_A} = n + (\phi-1)g_A.$$

Multiply both sides of the equation by $g_A$:

$$ \dot{g_A} = n g_A + (\phi-1) g_A^2$$

Given that $\phi&lt;1$, and in the steady-state BGP the growth of the growth rate $\dot{g_A} = 0$, we divide both sides by $g_A$ to get:

$$n+ (\phi -1) g_A = 0 $$ 

Rearrange to get the desired expression:

$$g_A^* = \frac{n}{1-\phi}$$

The second equation is just the result above multiplied by $\sigma$:

$$g_Y = g_A \sigma =  \frac{\sigma n}{1-\phi}$$