How to find the rate of a regression equation?

In all fairness GoldenLeaf you have answered this question very well.

HOWEVER

The question that the asker intended is not the one that you answered. 

The intended question was 

6872=14592(1+r/1)^11

I know this because the word exponential was used in the question.

$$6872=14592(1+r)^{11}\\\\ \frac{6872}{14592}=(1+r)^{11}\\\\ log\left(\frac{6872}{14592}\right)=log(1+r)^{11}\\\\ log\left(\frac{6872}{14592}\right)=11log(1+r)\\\\ \dfrac{log\left(\frac{6872}{14592}\right)}{11}=log(1+r)\\\\ 10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}=10^{log(1+r)}\\\\ 10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}=1+r\\\\ 10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}-1=r\\\\$$

$${{\mathtt{10}}}^{\left({\frac{{log}_{10}\left({\frac{{\mathtt{6\,872}}}{{\mathtt{14\,592}}}}\right)}{{\mathtt{11}}}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.066\: \!165\: \!642\: \!977\: \!294\: \!8}}$$

$$r\approx -0.066$$

Just plug it in!

$${\mathtt{6\,872}} = {\mathtt{14\,592}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{r}}}{{\mathtt{1}}}}\right){\mathtt{\,\times\,}}{\mathtt{11}} \Rightarrow {\mathtt{r}} = {\mathtt{\,-\,}}{\frac{{\mathtt{19\,205}}}{{\mathtt{20\,064}}}} \Rightarrow {\mathtt{r}} = -{\mathtt{0.957\: \!187\: \!001\: \!594\: \!896\: \!3}}$$

I'll try to show some work, but gimme a sec.

As far as I know, we need to isolate the 'r'.

First combine like terms.

$${\mathtt{6\,872}} = {\mathtt{160\,512}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{r}}}{{\mathtt{1}}}}\right)$$

Divide 6872 by 160512

$${\frac{{\mathtt{6\,872}}}{{\mathtt{160\,512}}}} = {\mathtt{0.042\: \!812\: \!998\: \!405\: \!103\: \!7}}$$

$${\mathtt{0.042\: \!812\: \!998\: \!405\: \!103\: \!7}} = {\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{r}}}{{\mathtt{1}}}}$$

The 'r/1' can be just 'r'. All that's needed is to subtract 1 from 0.0428129984051037

$${\mathtt{0.042\: \!812\: \!998\: \!405\: \!103\: \!7}}{\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.957\: \!187\: \!001\: \!594\: \!896\: \!3}}$$

$$-{\mathtt{0.957\: \!187\: \!001\: \!594\: \!896\: \!3}} = {\mathtt{r}}$$

And thus, the work =)