# Christina put 15000 euro in the bank with a monthly interest rate of .6%. When can she buy a car that costs 24800 euro and devaluates.8% per month

Christina has $15000$ euro and wants to buy a car. Presently, this car costs $24800$ euro. Christina will put the $15000$ euro in the bank with a monthly interest rate of $.6\%$. The car devaluates $.8\%$ a month.

How many months will it take until Christina is able to buy the car?

I did:

$$15000 \cdot (1+.6)^t = 24800 \cdot (1-.8)^t \Leftrightarrow \frac{15000}{24800} = \frac{.2^t}{1.6^t} \Leftrightarrow \frac{75}{124} = (\frac{.2}{1.6})^t \Leftrightarrow \frac{75}{124} = (\frac{5}{40})^t \Leftrightarrow \frac{75}{124} = (\frac{1}{8})^t \Leftrightarrow \log_{\frac{1}{8}}(\frac{75}{124})=t \Leftrightarrow t \approx .24179...$$

But this result doesn't make any sense. My book says the solution is 36 months.

What did I do wrong?

• The $.6$ and $.8$ should be $0.006$ and $0.008$ because they are percentages. – AndroidFish Nov 25 '16 at 3:42

$.6\%$ is $.006$. Do likewise for $.8\%$. Your error comes from using $.6\%$ as $.6$, which equals $60\%$.