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

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))

• Please show what you obtained by applying the logarithm rules. Someone may be able to identify a simple algebraic or conceptual error, or suggest a next step, without having to duplicate your effort. ... Since comments are easily overlooked, please edit your question to include such details.
– Blue
Dec 15 '19 at 16:57
• Hint: $$\log\frac x{1+\beta}=\log x - \log(1+\beta)$$ and $$\log(x\beta)=\log\beta+\log x$$ after doing that you only have to collect the terms Dec 15 '19 at 16:57
• @MaximilianJanisch: Need one more hint. I have expanded using your hint. Dec 15 '19 at 17:14
• @user508281 what do you have now? Dec 15 '19 at 17:14
• I have edited the original question. Check equation 4. Dec 15 '19 at 17:14

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