Please help me, I have to solve these problems:
My great-grandfather invested $ 9.00 in a savings account in 1909. The amount of money in the account double every 9 years.
* How much money would be in the account 99 years later?-------------
* By what percentage does the investment grow in one year?---------
* What year will it be when the value of his investment hits $ 1 million?----
1909 = 9
1918 = 18
1927 = 36
1936 = 72
1945 = 144
1954 = 288
1963 = 576
Take it from there.
Sure! I can help you solve these problems step by step.
1. To find out how much money would be in the account 99 years later, we need to determine how many times the initial amount doubles over that period of time. Since the amount doubles every 9 years, we divide 99 by 9 to get the number of times it doubles:
99 years / 9 years = 11 times
Now, we can calculate the final amount by multiplying the initial amount ($9.00) by 2 raised to the power of the number of times it doubles (11):
Final amount = $9.00 * 2^11
Final amount = $9.00 * 2048
Final amount = $18,432.00
Therefore, after 99 years, there would be $18,432.00 in the account.
2. To calculate the percentage by which the investment grows in one year, we can use the formula:
Percentage growth = (Final amount - Initial amount) / Initial amount * 100
In this case, the final amount is the amount after 9 years since the investment doubles in that period. We can use the formula to calculate the percentage growth:
Percentage growth = ($9.00 * 2 - $9.00) / $9.00 * 100
Percentage growth = $9.00 / $9.00 * 100
Percentage growth = 100%
The investment grows by 100% in one year.
3. To find out the year in which the value of the investment hits $1 million, we can use the formula:
Final amount = Initial amount * (2^(number of times it doubles))
We need to solve this formula for the number of times it doubles. Since we know the final amount ($1 million), the initial amount ($9.00), and the doubling period (9 years), we can rearrange the formula:
1 million = $9.00 * 2^(number of times it doubles)
Now, we need to solve for the number of times it doubles. We can use logarithms to do that:
log (base 2) of 1 million = log (base 2) of ($9.00 * 2^(number of times it doubles))
log (base 2) of 1 million = log (base 2) of $9.00 + log (base 2) of 2^(number of times it doubles)
The logarithm of 1 million to the base 2 is the same as asking, how many times do we need to double 1 to get 1 million:
log (base 2) of 1 million ≈ 19.93
Now, we can solve for the number of times it doubles:
19.93 = log (base 2) of $9.00 + number of times it doubles * log (base 2) of 2
19.93 ≈ 0.954 + number of times it doubles
Subtracting 0.954 from both sides:
number of times it doubles ≈ 19.93 - 0.954
number of times it doubles ≈ 18.976
Since each doubling takes 9 years, we can calculate the number of years it takes for the value of the investment to hit $1 million:
Number of years = number of times it doubles * 9
Number of years ≈ 18.976 * 9
Number of years ≈ 170.784
Therefore, it will be approximately 171 years when the value of his investment hits $1 million.
I hope this explanation helps! Let me know if you have any further questions.