1. HELP!!!Word Problem

If your allowance was a penny doubled every day, how many days would it take for you to have over $1000? On which day would you have over$10000? How much money would you have on the 15th day?

2. y = (1+r)^2

y = (1+200%)^2

y = (1+2) ^2

y = 3^2

y = 9

So, on the fifteenth day, you will have $9 This is an example of exponential growth (y = x(1+r)^t where x = initial starting amount, r = rate of growth expressed as a percent and t = time passed under specific unit). Plugging in your other numbvers ($1000 and $10000) as y will give you the answer to the other problems. ****Note that I may be wrong and it may be$6.25. I'm pretty sure that $9 is right, but for some reason my instinct told me to go with 150% the first time I went through with this. Although 150% is NOT doubled. Please check this with your own work or through someone else.**** 3. ? The answer is day 18$1308.16, day 21 $10465.28, on the 15th day$163.52.

4. Originally Posted by learnmath
The answer is day 18 $1308.16, day 21$10465.28, on the 15th day $163.52. I don't know how to calculate this. Please help!!! Day 1:$0.01

Day 2: $2(0.01) Day 3: 2(2(0.01)) =$\displaystyle \$2^2 (0.01)$

Day 4: $\displaystyle 2 (2^2 (0.01)) = \$ 2^3 (0.01)$etc. Day n:$\displaystyle \$2^{n-1} (0.01)$.

Q1 Solve $\displaystyle 2^{n-1} (0.01) > 1000 \Rightarrow 2^{n-1} > 100,000$.

The are many ways of solving this. Using logarithms is one way:

Spoiler:
$\displaystyle \log_{10} 2^{n-1} > \log_{10} 100,000 \Rightarrow (n-1) \log_{10} 2 > 5 \Rightarrow n - 1 > \frac{5}{\log_{10} 2} \Rightarrow n - 1 > 16.6$

Q2 Day 15:
Spoiler:
$\displaystyle \$ 2^{15-1} (0.01)\$