Proving a result In Jones (1999) "Growth: With or Without Scale Effects" Jones (1999) builds his semi-endogenous growth model where output is produced with only one input, labour, in the following "research" function:

$$ Y = A^\alpha L_Y $$

Labour is augmented with technology $A$ which is also researched by
workers, $L_A$, in the following production function:

$$ \dot{A} = \delta L_A A^\phi. $$

We have that $L_A + L_Y = L$ and $0 &lt; \phi &lt; 1$.

He then writes:


  Assuming that the labour force L grows at some constant exogenous rate
  $n$, it is easy to show that there exists a stable balanced growth
  path for the model where
  
  $$ g_A = \frac{n}{1-\phi}$$
  
  and
  
  $$ g_Y = \sigma g_A = \frac{\sigma n}{1-\phi} $$


He says it's "easy to show", but I can't for the life of me show it! We know that $g_A = \dot{A}/A$, and on the balanced growth path $g_A = g_Y = g_L = n$. So I try to plug $g_A = \delta L_A A^{\phi-1} = n$ but that gets me nowhere.
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 $\sigma$:

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