Answer:
It cost $0.91 10 years ago.
It takes 10.24 years for the cost of bread to double.
Step-by-step explanation:
The equation for the price of bread after t years has the following format:
![P(t) = P(0)(1+r)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29%281%2Br%29%5E%7Bt%7D)
In which P(0) is the current price, and r is the inflation rate, as a decimal.
If we want to find the price for example, 10 years ago, we find P(-10).
Inflation is at a rate of 7% per year. Evan's favorite bread now costs $1.79.
This means that
. So
![P(t) = P(0)(1+r)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29%281%2Br%29%5E%7Bt%7D)
![P(t) = 1.79(1+0.07)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%201.79%281%2B0.07%29%5E%7Bt%7D)
![P(t) = 1.79(1.07)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%201.79%281.07%29%5E%7Bt%7D)
What did it cost 10 years ago?
![P(-10) = 1.79(1.07)^{-10} = 0.91](https://tex.z-dn.net/?f=P%28-10%29%20%3D%201.79%281.07%29%5E%7B-10%7D%20%3D%200.91)
It cost $0.91 10 years ago.
How long before the cost of the bread doubles?
This is t for which P(t) = 2P(0) = 2*1.79. So
![P(t) = 1.79(1.07)^{t}](https://tex.z-dn.net/?f=P%28t%29%20%3D%201.79%281.07%29%5E%7Bt%7D)
![2*1.79 = 1.79(1.07)^{t}](https://tex.z-dn.net/?f=2%2A1.79%20%3D%201.79%281.07%29%5E%7Bt%7D)
![(1.07)^{t} = 2](https://tex.z-dn.net/?f=%281.07%29%5E%7Bt%7D%20%3D%202)
![\log{(1.07)^{t}} = \log{2}](https://tex.z-dn.net/?f=%5Clog%7B%281.07%29%5E%7Bt%7D%7D%20%3D%20%5Clog%7B2%7D)
![t\log{1.07} = \log{2}](https://tex.z-dn.net/?f=t%5Clog%7B1.07%7D%20%3D%20%5Clog%7B2%7D)
![t = \frac{\log{2}}{\log{1.07}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Clog%7B2%7D%7D%7B%5Clog%7B1.07%7D%7D)
![t = 10.24](https://tex.z-dn.net/?f=t%20%3D%2010.24)
It takes 10.24 years for the cost of bread to double.