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An investment earning interest at the rate of [tex]10\%[/tex], compounded continuously, will double in [tex]t[/tex] years. Find [tex]t[/tex].

Use the formula [tex]P(t) = P_0 e^{rt}[/tex], where [tex]P(t)[/tex] is the amount after [tex]t[/tex] years, [tex]P_0[/tex] is the initial amount, [tex]r[/tex] is the rate of interest, and [tex]t[/tex] is the time.

[tex]\boxed{\text{ } \text{ years}}[/tex]

Sagot :

GinnyAnswer
To determine the time [tex]\( t \)[/tex] required for an investment to double when it is earning interest at a rate of 10%, compounded continuously, we can use the formula for continuous compounding:

[tex]\[ P_t = P_0 e^{rt} \][/tex]

Here:
- [tex]\( P_t \)[/tex] is the amount after [tex]\( t \)[/tex] years.
- [tex]\( P_0 \)[/tex] is the initial amount.
- [tex]\( r \)[/tex] is the interest rate.
- [tex]\( t \)[/tex] is the time in years.
- [tex]\( e \)[/tex] is the base of the natural logarithm.

Our goal is to find the time [tex]\( t \)[/tex] when the investment doubles, which means [tex]\( P_t = 2P_0 \)[/tex]. We substitute [tex]\( P_t \)[/tex] and [tex]\( r \)[/tex] into the formula:

[tex]\[ 2P_0 = P_0 e^{0.10t} \][/tex]

We can divide both sides by [tex]\( P_0 \)[/tex] to simplify:

[tex]\[ 2 = e^{0.10t} \][/tex]

To solve for [tex]\( t \)[/tex], we take the natural logarithm (ln) of both sides:

[tex]\[ \ln(2) = \ln(e^{0.10t}) \][/tex]

Using the property of logarithms that [tex]\( \ln(e^x) = x \)[/tex]:

[tex]\[ \ln(2) = 0.10t \][/tex]

Now, solve for [tex]\( t \)[/tex]:

[tex]\[ t = \frac{\ln(2)}{0.10} \][/tex]

The value of [tex]\( \ln(2) \)[/tex] is approximately 0.693. Therefore:

[tex]\[ t = \frac{0.693}{0.10} \approx 6.931 \][/tex]

Rounding this value to the nearest whole number gives:

[tex]\[ t \approx 7 \][/tex]

So, the time it takes for the investment to double is approximately

[tex]\[ \boxed{7} \text{ years} \][/tex]
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