The function #f(t)=Pe^(rt)# describes the amount #P# will become if invested for #t# years at #r%# per annum compounded continuously. If a sum becomes #$246.40# at #4%# in #4# years, what is the amount invested?

Question

Eva Adamson · Accepted Answer

#P~=210# As we have the function #f(t)=Pe^(rt)# and #f(4)=246.4# we have #246.4=Pe^(0.04xx4)=Pe^(0.16)# Hence #P=246.4/e^(0.16)=246.4/1.173510871=209.968~=210#

Genesis Allen · Answer

undefined To find the amount invested (P), we can use the formula $ P = \frac{A}{e^{rt}} $, where A is the final amount, r is the interest rate (in decimal form), t is the time in years, and e is the base of the natural logarithm (approximately equal to 2.71828).

Given:
A = $246.40
r = 4% = 0.04 (in decimal)
t = 4 years

Using the given values in the formula, we get:
$$ P = \frac{246.40}{e^{0.04 \times 4}} $$

$$ P \approx \frac{246.40}{e^{0.16}} $$

$$ P \approx \frac{246.40}{1.1735} $$

$$ P \approx 209.88 $$

So, the amount invested is approximately $209.88.

Julia Adams · Answer

undefined To find the amount invested, we can use the formula for continuous compounding: $ P = \frac{A}{e^{rt}} $, where $ P $ is the initial principal, $ A $ is the final amount, $ r $ is the interest rate, and $ t $ is the time in years.

Given:
$ A = 246.40 $
$ r = 0.04 $ (4% expressed as a decimal)
$ t = 4 $

We can rearrange the formula to solve for $ P $:
$ P = \frac{A}{e^{rt}} $

Substituting the given values:
$ P = \frac{246.40}{e^{0.04*4}} $

Now, calculate:
$ P = \frac{246.40}{e^{0.16}} $

Using the approximate value of $ e $ (2.71828):
$ P = \frac{246.40}{2.71828^{0.16}} $

Calculate $ e^{0.16} $:
$ e^{0.16} ≈ 1.17351087 $

Now divide:
$ P ≈ \frac{246.40}{1.17351087} $

$ P ≈ 210.00 $

So, the amount invested is approximately $210.