1. If you invested money into an account that pays 9% compounded weekly, how many years would it take for your deposit to double?
We're in the unit "Exponential and Logarithmic Functions"
I would really appreciate it if you could help me! :D
This is what I did:
y=a(1+b)^n
where b is percentage, n is variable for time, a is initial value and y is final value.
I decided to plug in values to see what the numbers would become.
200=100(1+0.09)^n
2=(1.09)^n
then from exponential to logarithm converted to
log[base 1.09] (2)=n
then
log2/log1.09=n
n=8.04
I figured if it was compounded weekly, it would be 8 weeks = 2 months
Then 2 months/12 months per year
0.16 years.
I might have done something wrong, Thanks!
We're in the unit "Exponential and Logarithmic Functions"
I would really appreciate it if you could help me! :D
This is what I did:
y=a(1+b)^n
where b is percentage, n is variable for time, a is initial value and y is final value.
I decided to plug in values to see what the numbers would become.
200=100(1+0.09)^n
2=(1.09)^n
then from exponential to logarithm converted to
log[base 1.09] (2)=n
then
log2/log1.09=n
n=8.04
I figured if it was compounded weekly, it would be 8 weeks = 2 months
Then 2 months/12 months per year
0.16 years.
I might have done something wrong, Thanks!
-
amount = principal(1 + 0.09/52)^52t . . . let the principal = 1
2 = (1 + 0.09/52)^52t
2 = (1.00173)^52t
In2 = 52t*In(1.00173) . . . taking natural log of both sides and using properties of logarithms
In2/In(1.00173) = 52t
401.00958 = 52t
t = 7.71 years
The quick method for continuous compounding is In2/rate = 7.7 years
2 = (1 + 0.09/52)^52t
2 = (1.00173)^52t
In2 = 52t*In(1.00173) . . . taking natural log of both sides and using properties of logarithms
In2/In(1.00173) = 52t
401.00958 = 52t
t = 7.71 years
The quick method for continuous compounding is In2/rate = 7.7 years