$100 is deposited in a savings account that pays 4% interest compounded annually.
A) how much is in the account after 5 years?
B) use logarithms to determine how many years it will take for the account to grow to $1000.
Could someone show me how to do this step by step? Thank you
a) amount = 100(1.04)^5
= ...
b)
100(1.04)^n = 1000
1.04^n = 10
log[ 1.04^n ] = log 10
n log 1.04 = 1
n = 1/log 1.04 = ...
(you do the button-pushing)
Sure! I'd be happy to walk you through the steps of solving these questions.
A) To find out how much money is in the account after 5 years, we will use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A = the final amount in the account
P = the initial principal amount (in this case $100)
r = annual interest rate (in decimal form, so 4% is 0.04)
n = number of times the interest is compounded per year (annually in this case, so n = 1)
t = number of years
Plugging in the values, the formula becomes:
A = 100(1 + 0.04/1)^(1 * 5)
A = 100(1.04)^(5)
A = 100(1.216653)
A ≈ 121.67
So, after 5 years, there will be approximately $121.67 in the savings account.
B) To determine how many years it will take for the account to grow to $1000 using logarithms, we can use the following formula:
t = log(base, (A/P)) / log(base, (1 + r/n))
In this case, we want to find the value of t. We know A ($1000), P ($100), r (0.04), and n (1). We need to find the appropriate base for the logarithm.
Now, we can rearrange the formula to solve for t:
t = [log(base, (A/P))] / [log(base, (1 + r/n))]
Since you mentioned using logarithms, let's use base 10 logarithm (log10) for simplicity.
t = [log10(1000/100)] / [log10(1 + 0.04/1)]
Using a calculator:
t ≈ log10(10) / log10(1.04)
t ≈ 1 / 0.0170
t ≈ 58.82
So, it will take approximately 58.82 years for the account to grow to $1000.
I hope this helps! Let me know if you have any further questions.