Figuring point in savings account

  • #1

Main Question or Discussion Point

say i have an amount of say $20000 in my savings account and at 2% interest per month when will I have 1 million?

my current method is 20000(1.02)x=1000000, and solve for x

I get 197, which is months, 197/12=16.4yr. about. is that right?
 

Answers and Replies

  • #2
98
0
I guess what you want is a simple compound interestformula

A = P ( 1 + r/n ) ^nt

Amount is equal to princple times 1 plus rate divided by number of times interest is compounded per year to the numbver of times interest is compounded per year times time.

PLug in your values and solve for t
 
  • #3
so 1000000=20000(1+0.02/12)nt what answer would you get?
 
  • #4
98
0
Not exactly, n would be 6 because your interest is counted every 2 months .

Also, you have to use the n with the exponent too. After that plug and chug and solve for t. You might have to use logs.
 
  • #5
thanks, but i dont know logs yet, im learning them in this unit tho. I was wondering if someone who did know would figure it for me.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
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Not exactly, n would be 6 because your interest is counted every 2 months .

Also, you have to use the n with the exponent too. After that plug and chug and solve for t. You might have to use logs.
Unless it was editted, I don't see where the original post said anything about "every 2 months".

so 1000000=20000(1+0.02/12)nt what answer would you get?
No, that would be for 2% per year and the question is about 2% per month (24% annual- PLEASE tell me where you can get that!). The formula in the orginal post was correct.

[itex]20000(1.02)^x=1000000[/itex]
so [itex]1.02^x= 1000000/20000= 50-[/itex], x log(1.02)= log(50),
x= log(50)/log(102)= 198 months, about 16 and a half years, as the original post said. What was all the fuss about?
 
  • #7
98
0
Unless it was editted, I don't see where the original post said anything about "every 2 months".


No, that would be for 2% per year and the question is about 2% per month (24% annual- PLEASE tell me where you can get that!). The formula in the orginal post was correct.

[itex]20000(1.02)^x=1000000[/itex]
so [itex]1.02^x= 1000000/20000= 50-[/itex], x log(1.02)= log(50),
x= log(50)/log(102)= 198 months, about 16 and a half years, as the original post said. What was all the fuss about?
ah my mistake, I put the 2 from the 2percent in front of the months. It happens.
 
  • #8
I was just seeing if my math was right, and its actually 4% annual, I had a bit of a mix up.
 

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