word problem: investments

kellyanne

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Aug 22, 2006
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determine how much time is required for an investment to double in value if interest is earned at a rate of 6.25% compounded continuously.
 
You're not showing much work on these.

Compound Annually \(\displaystyle (1 + 0.0625)^{x} = 2\)

Compound Semi-Annually \(\displaystyle [(1 + 0.0625/2)^{2}]^{x} = 2\)

Compound Quarterly \(\displaystyle [(1 + 0.0625/4)^{4}]^{x} = 2\)

Compound Monthly \(\displaystyle [(1 + 0.0625/12)^{12}]^{x} = 2\)

Compound Weekly \(\displaystyle [(1 + 0.0625/52)^{52}]^{x} = 2\)

Compound Daily \(\displaystyle [(1 + 0.0625/365)^{365}]^{x} = 2\)

Compound Hourly \(\displaystyle [(1 + 0.0625/(365*24))^{365*24}]^{x} = 2\)

Compound Minutely \(\displaystyle [(1 + 0.0625/(365*24*60))^{365*24*60}]^{x} = 2\)

What's the limit of this process?
 
kellyanne said:
determine how much time is required for an investment to double in value if interest is earned at a rate of 6.25% compounded continuously.
Pick a variable for the original investment amount. What then is the expression for the doubled amount?

Plug the variable, the expression for the doubled amount, and the given interest rate into the continous-compounding formula you've memorized. Solve for the time t.

Eliz.
 
tkhunny said:
Compound Daily \(\displaystyle [(1 + 0.0625/365)^{365}]^{x} = 2\)
Compound Hourly \(\displaystyle [(1 + 0.0625/(365*24))^{365*24}]^{x} = 2\)
Compound Minutely \(\displaystyle [(1 + 0.0625/(365*24*60))^{365*24*60}]^{x} = 2\)

Them 365's should be 365.25 :idea:
 
Denis said:
Them 365's should be 365.25
Or they could be 360's, if one uses the Banker's Rule. It depends upon how the "year" is defined. In most pre-calc algebra courses, in my experience, a "year" has 365 days.

Eliz.
 
Denis said:
Them 365's should be 365.25 :idea:
I tend to lean toward 365.2425, myself, but generally leave that to the folks who will live long enough to care. :)

Thanks for the chuckle.
 
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