How much does each fraction contribute to decreasing a value? Say I have a value: 0.68. And this is decreased in half to 0.34 by multiplying it by $\frac{3}{4}$ and $\frac{2}{3}$. Let's consider this decrease of $\frac{1}{2}$ as 100% of the decrease. So then we can consider the idea that $\frac{3}{4}$ and $\frac{2}{3}$ each contributed to this decrease, but in different magnitudes. I want to quantify the magnitude that each contributed to the decrease by assigning a percent.
If the total product contributes 100% to decreasing the value from 0.68 to 0.34, how can I get two percentages that represent how much the $\frac{3}{4}$ contributed and how much the $\frac{2}{3}$ contributed, respectively. Clearly the $\frac{2}{3}$ contributes more to the decrease of the value, but how much more?
All of my attempts have been squandered by, what I think is an error in, order of operations. If I pull out $\frac{3}{4}$ and $\frac{2}{3}$ and multiply the 0.68 individually, it doesn't make sense to pull it back in to see it's impact because I have just ruined my order of operations.
 A: The amount in the decrease in a value $n$ by multiplying by $\frac{3}{4}$ given by:
$$n\bigg(1-\frac{3}{4}\bigg)=n-\frac{3n}{4}= \frac{n}{4}$$
After this decrease, remains a value of $\frac{3n}{4}$.  Multiplying now by $\frac{2}{3}$ decreases this remaining value by:
$$\bigg(\frac{3n}{4}\bigg)\bigg(1-\frac{2}{3}\bigg)=\frac{3n}{4}-\frac{6n}{12}=\frac{9n-6n}{12}=\frac{3n}{12}=\frac{n}{4}$$
Therefore, they both decrease the value by the same amount, which makes sense.  If you have $n$ portions and you remove $\frac{1}{n}$ of it, you end up with ${n-1}$ portions.  So, you can successively multiply by $\frac{n-1}{n}$ with decreasing $n$ and remove the same amount each time.
A: Since we have
$$0.68\cdot\frac{2}{3}\cdot \frac{3}{4}=\frac12\cdot0.68=0.34$$
we should evaluate the contribute of $\frac{3}{4}$ and $\frac{2}{3}$ for $\frac12$ but the two moltiplicative coefficient are correlated and act together. 
To better see that fact we can consider
$$\frac{2}{3}\cdot\frac{3}{4}=\left(1-\frac13\right)\left(1-\frac14\right)=1-\frac13-\frac14+\frac1{12}$$
thus
$$0.68\cdot\frac{2}{3} \cdot\frac{3}{4}=0.68\cdot\left(1-\frac13\right)\left(1-\frac14\right)=0.68-\frac13\cdot 0.68-\frac14\cdot 0.68+\frac1{12}\cdot 0.68$$
then


*

*the reduction for $\frac23$ as it was applied singularly is$-\frac13\cdot 0.68=-0.22\bar6$

*the reduction for $\frac34$ as it was applied singularly is $-\frac14\cdot 0.68=-0.170$

*the increment for $\frac23$ and $\frac34$ acting together is$\frac1{12}\cdot 0.68=0.05\bar6$


More in general, if we indicate with $A_0$ the initial amount, with $A_f$ the final amount, with $\Delta A=A_0-A_f$ the variation and with $r_1=1-c_1$ and $r_2=1-c_2$ the reduction factors we have
$$A_f= A_0\cdot r_1 \cdot r_2= A_0\cdot (1-c_1) \cdot (1-c_2)=A_0-c_1A_0-c_2A_0+c_1c_2A_0\\\implies \Delta A=A_0-A_f=A_0(c_1+c_2-c_1c_2)\\\implies \frac{\Delta A}{A_0}=c_1+c_2-c_1c_2$$
As suggested and explained by JMoravitz, by logarithms it is possible to separate the two contributes and assign a petange to each one separetely.
