How to find the rate (r) of a exponential regression equation?
6872=14592(1+r/1)11
+118677 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.
$$6872=14592(1+r)^{11}\\\\
\frac{6872}{14592}=(1+r)^{11}\\\\
log\left(\frac{6872}{14592}\right)=log(1+r)^{11}\\\\
log\left(\frac{6872}{14592}\right)=11log(1+r)\\\\
\dfrac{log\left(\frac{6872}{14592}\right)}{11}=log(1+r)\\\\
10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}=10^{log(1+r)}\\\\
10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}=1+r\\\\
10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}-1=r\\\\$$
$${{\mathtt{10}}}^{\left({\frac{{log}_{10}\left({\frac{{\mathtt{6\,872}}}{{\mathtt{14\,592}}}}\right)}{{\mathtt{11}}}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.066\: \!165\: \!642\: \!977\: \!294\: \!8}}$$
$$r\approx -0.066$$
+1006 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 =)
+118677 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.
$$6872=14592(1+r)^{11}\\\\
\frac{6872}{14592}=(1+r)^{11}\\\\
log\left(\frac{6872}{14592}\right)=log(1+r)^{11}\\\\
log\left(\frac{6872}{14592}\right)=11log(1+r)\\\\
\dfrac{log\left(\frac{6872}{14592}\right)}{11}=log(1+r)\\\\
10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}=10^{log(1+r)}\\\\
10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}=1+r\\\\
10^{\frac{log\left(\frac{6872}{14592}\right)}{11}}-1=r\\\\$$
$${{\mathtt{10}}}^{\left({\frac{{log}_{10}\left({\frac{{\mathtt{6\,872}}}{{\mathtt{14\,592}}}}\right)}{{\mathtt{11}}}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.066\: \!165\: \!642\: \!977\: \!294\: \!8}}$$
$$r\approx -0.066$$
