How Long to Double Your Money with Compound Interest?

In summary, to find how long it will take for $30 000 to double in value with an interest rate of 4.3% compounded annually, we use the formula S(1+ r)^n = 2S, where S is the initial amount and r is the interest rate. After solving for n, we get n = log2 / log(1+r), which in this case is approximately 16.22 years. This means it will take about 16.22 years for $30 000 to double in value under these conditions.
  • #1
jahaddow
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A fixed deposit investment attracts interest of 4.3% p.a. compounded annually. How long will $30 000 take to double in value?
 
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  • #2
jahaddow said:
A fixed deposit investment attracts interest of 4.3% p.a. compounded annually. How long will $30 000 take to double in value?

What have you tried? What are the relevant equations?
 
  • #3
This is all I know, I don't know what method should be used to find the answer?
 
  • #4
That's very peculiar! If you aren't taking a course that has discussed "compound interest", why have you been assigned a problem like this? If you are then surely there are formulas for compound interest in your textbook aren't there?

If I remember correctly, after n years, an amount S at r rate of interest (so that 100r% is the annual percentage rate) compounded annually is given by [itex]S(1+ r)^n[/itex]. Set that equal to 2S, cancel the "S"s and solve for n.
 
  • #5
A powerful strategy is to think about the specific case and try to obtain a general equation.

The first year, you have $30 000

The second year it is compounded by $30 000 x .043 so you have 30 000 + 30 000 x .043
= 30 000 (1.043)

The next year your money increases again by 30 000 (1.043) (.043) for a total of 30 000 (1.043) + 30 000 (1.043) (.043) = 30 000 (1.043) (1.043) = 30 000 (1.043)^2

By now you should have noticed a pattern.
 

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