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I WILL GIVE BRAINLIEST Suppose that is $1200 initially invested in an account at a fixed interest rate, compounded continuously. Suppose also that, after six years, the amount of money in the account is $1431 . Find the interest rate per year.
Write your answer as a percentage. Do not round any intermediate computations, and round your percentage to the nearest hundredth.

I WILL GIVE BRAINLIEST Suppose That Is 1200 Initially Invested In An Account At A Fixed Interest Rate Compounded Continuously Suppose Also That After Six Years class=

Sagot :

Answer:  2.93

Note: The percent sign is already taken care of.

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Explanation:

The continuously compounded interest formula is

[tex]A = Pe^{rt}\\\\[/tex]

where,

  • A = 1431 = account balance t years into the future
  • P = 1200 = deposit
  • e = special constant roughly equal to 2.718
  • r = unknown interest rate and what we want to find
  • t = 6 years

Since the unknown variable r is in the exponent, we'll need logs to solve it. A handy phrase is "if the exponent is in the trees, log it down to find it". By "log it down" I mean in terms of lumber, and not a journal.

We could use any base logarithm, but it's handy to use the special base 'e' log. This is the natural logarithm abbreviated as Ln.

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Solving for r involves these steps

[tex]A = Pe^{rt}\\\\1431 = 1200*e^{r*6}\\\\e^{6r} = 1431/1200\\\\e^{6r} = 1.1925\\\\6r = \text{Ln}(1.1925)\\\\r = \frac{1}{6}*\text{Ln}(1.1925)\\\\r \approx 0.02934199063006\\\\r \approx 0.0293\\\\[/tex]

Move the decimal over 2 spots to the right to convert 0.0293 to 2.93%

This is the same as multiplying by 100.

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