Reducing $\log\frac{x}{1+\beta}+\beta\log\left(x-\frac{x}{1+\beta}\right)$

Question

I am trying to verify that the expression in line 1 boils down to the expression in line 3.

From line 1 to line 2, it is simple.

However, I don't get how the final expression in line 3 is derived.

I have tried using the quotient rule (Log simplification rule) but could not get the last expression.

I'd really appreciate if anyone could guide me where to start.

$$\log\frac{x}{1+\beta}+\beta\log\left(x-\frac{x}{1+\beta}\right) \tag{1}$$ $$\log\frac{x}{1+\beta}+\beta\log\frac{\beta x}{1+\beta} \tag{2}$$ $$(1+\beta)\log x + \beta\log\beta-(1+\beta)\log(1+\beta) \tag{3}$$

I get:

$$(\log x -\log(1+\beta) +\beta(\log\beta+\log x-\log(1+\beta)) \tag{4}$$

(original problem image (the above replaces $X_{T-1}$ with $x$ to reduce visual clutter))

Answer 1

Hint:

Write $\;\beta\log\dfrac{\beta x}{1+\beta}=\beta\log\Bigl(\beta\,\dfrac{x}{1+\beta}\Bigr)=\beta\log\beta+\beta\log\dfrac{x}{1+\beta}$, whence $$\log\frac{x}{1+\beta}+\log\Bigl(\beta\frac{x}{1+\beta}\Bigr)=(1+\beta)\log\frac{x}{1+\beta}+\beta\log\beta.$$

Answer 2

$$log \frac{x}{1+\beta}+\beta log \frac{\beta x}{1+\beta}=\log\frac{x}{1+\beta} \big[ \big( \frac {x}{1+\beta}\big )^{\beta} +\beta^{\beta}\big]= log \big[ \big( \frac{x}{1+\beta}\big)^{\beta+1}+\big(\frac{\beta^{\beta} x}{1+\beta}\big)\big]= log \frac{x+(1+\beta)^{\beta}\beta^{\beta }x}{(1+\beta)^{1+\beta}}=log\big(\frac{\beta^{2\beta}+\beta^{\beta}+1}{(1+\beta)^{\beta +1}}\big) x$$