To simplistically answer your question, use the following:
$Y = K^\alpha (AL)^{1 - \alpha}$
In order to prove all three can be equal:
We will assume that technological progress is labor augmenting (Harrod Neutral) or "labor saving", while holding the following true:
- Returns to capital are roughly constant over time.
- Capital share of income is roughly constant over time.
The Harrod Neutral is associated with a capital-output ratio that is constant (K/Y) in our "steady state".
Capital Accumulation
In the basic Solow model, (1).
$$ \frac{\dot{k}}{k} \equiv s\frac{y}{k}-(d+n)$$
For growth rate per capita capital to be constant, y/k = Y/K must also be constant. (Characteristic of the Harrod Neutral)
We Need To Define The Balanced Growth Path With Technological Progress
Taking the original production equation:
$Y = K^\alpha (AL)^{1 - \alpha}$
and dividing by AL (number of effective labor units) we are given our production function representing effective labor.
$$ \tilde{y} \equiv \tilde{k}^\alpha$$
Doing The Same For Capital Accumulation
Follow steps below:
$$\frac{\tilde{\dot{k}}}{k} = \frac{dln(k/AL)}{dt}$$
$$ \frac{dlnK}{dt} - \frac{dlnA}{dt} - \frac{dlnL}{dt} $$
$$\frac{\dot{K}}{K} - \frac{\dot{A}}{A} - \frac{\dot{L}}{L} = \frac{\dot{K}}{K} - g - n$$
then using this to finding our Capital Accumulation equation below:
$$\frac{\dot{K}}{K} = S \frac{{Y}}{K} - d$$
$$\frac{\tilde{\dot{k}}}{\tilde{k}} = s\frac{Y}{K} - d - g - n$$
And, $$\frac{Y}{K} = \frac{Y}{AL} \frac{AL}{K} = \frac{\tilde{y}}{\tilde{k}}$$
Then,
$$\frac{\tilde{\dot{k}}}{\tilde{k}} = s\frac{\tilde{y}}{\tilde{k}} - d - g - n$$
$$\tilde{\dot{k}} = s \tilde{y} - (d+g+n)\tilde{k}$$
Finally, combining our production function with capital accumulation we get an equation (6) that represents the Solow model with tech changes.
$$\tilde{\dot{k}} = s \tilde{k^\alpha} - (d+g+n)\tilde{k}$$
So our steady state is when (7) is true.
$$\tilde{\dot{k}^*} = 0$$
If we set this to be true for our steady state (denoted by *) levels of capital and output per effective labor. (8 and 9)
8: $$\tilde{k^*} = [\frac{s}{d+n+g}]^\frac{1}{1-\alpha}$$
9: $$\tilde{y^*} = [\frac{s}{d+n+g}]^\frac{\alpha}{1-\alpha}$$
In the steady state: (10)
Capital (and income) per capita grow at the rate of exogenous technological growth.
$$\tilde{k} = K/AL$$
$$k = K/L$$
$$k = A\tilde{k}$$
$$\tilde{k} = \tilde{k}^* = [\frac{s}{d+n+g}]^\frac{1}{1-\alpha}$$
$$k(t) = A(t)\tilde{k}^*$$
$$lnk(t) = lnA(t) + ln\tilde{k}^*$$
$$\frac{\dot{k}}{k} = \frac{\dot{A}}{A}$$
If you graph out
$$\tilde{\dot{k}} = s \tilde{k^\alpha} - (d+g+n)\tilde{k}$$
- Any savings rate changes, s, change the savings curve and therefore change $$\tilde{k^*}$$
- Population or depreciation changes, n and d respectively, rotates the straight line, (d+g+n)k.
Changes in these affect the growth rate of per capita capital and per capita income only outside the steady-state.
- exo-tech growth changes the graph sort of like n and d by changing the level of $$\tilde{k^*}$$
This exo-tech growth changes the growth rate of per capita capital and per capita income in the steady state (along balanced growth paths).
