Find time for investment to double, given interest rate

[manuel1]

Po is invested in a savings account in which interest is compounded continuously at 5.2% pear year. That is, the balance P grows at the rate given by dP/dt = 5.2P. Suppose that $6000 is invested. When will the investment double?

Thank you

 

[tkhunny]

Perhaps you have not yet seen my challenge in your previous post. Please read it before changing names again.

 

[stapel]

Po is invested in a savings account in which interest is compounded continuously at 5.2% pear year. That is, the balance P grows at the rate given by dP/dt = 5.2P. Suppose that $6000 is invested. When will the investment double?

Try using what you learned back in algebra, using the continously-compounded interest formula. Use logs to solve. :wink:

Eliz.

 

[ladyazpy]

Po= e^(052t)
Po = 6000

2000= 6000 e^(052t)

3= e^(052t)

ln3 = ln e^(052t)

ln 3 = (052t)

ln3/(052t) = t

t= 4.06

so does it means that the investment of $6000 will be doubled in 4 years and 1 month

 

[skeeter]

Po= e^(052t)
Po = 6000

2000= 6000 e^(052t) ???

3= e^(052t)

ln3 = ln e^(052t)

ln 3 = (052t)

ln3/(052t) = t

t= 4.06

so does it means that the investment of $6000 will be doubled in 4 years and 1 month

double the investment in 4 years ... ??? I don't think so.

P = P_oe^.052t

12000 is double the initial investment ...

12000 = 6000e^.052t

2 = e^.052t

ln(2) = .052t

t = ln(2)/.052 = 13.3 years

 

[stapel]

Po= e^(052t)

I will guess that you mean the exponent to be "0.052", since "052" isn't actually a number.

2000= 6000 e^(052t)

Actually, 2000 is not two times as much as 6000. Try the multiplication again: (2)(6) = ...? So (2)(6000) = ...? :wink:

3= e^(052t)

Actually, 2 / 6 is not 3, and "3" does not indicate "doubled". Hint: "Double" means "twice as much", and "twice" means "two times". :!:

ln3/(052t) = t

I will guess that you actually divided through by 0.052, leaving the "t" by itself, so you actually mean "ln(3) / 0.052 = t". :?:

t= 4.06

To check the solution to any "solving" problem, plug it back into the original exercise:

. . . . .A = P₀ e^rt

. . . . .A = 6000 e^{(0.052)(4.06)}

. . . . .A = 6000 e^0.21112

. . . . .A = (6000)(1.2350605534...)

. . . . .A = 7410.3633206....

Since $7410 is not twice as much as six thousand, then "t = 4.06" cannot be the solution. :shock:

Thank you!

Eliz.

 

[TchrWill]

Po is invested in a savings account in which interest is compounded continuously at 5.2% pear year. That is, the balance P grows at the rate given by dP/dt = 5.2P. Suppose that $6000 is invested. When will the investment double?

S = P(1 + i)^n

12000 = 6000(1 + .052/365)^n

2 = 1.000142465^n

log2 = log1.000142465n

n = 4865.7days = 13.33 years.

An old rule of thumb is that an amount, P,earning an annual interest rate of I,will approximately double in 72/I,or 72 divided by the whole interest rate, or in this case 72/5.2 = 13.84 years. Close.