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Answer:

Step-by-step explanation:

n is the number of times the interest compounds per year.  If the interest in this problem only compounds once per year ("annually"), then n = 1 and you'd be just as well off to use the formula:

[tex]A=P(1+r)^t[/tex]

When n = 1, r/n is just r.  But I'll show you using the formula they want you to use; it's the same anyways.

For us, P = 500, r = .015, n = 1.  Filling that into the formula:

[tex]A=500(1+\frac{.015}{1})^{(1)(t)}[/tex] which simplifies down to

[tex]A=500(1+.015)^t[/tex] and

[tex]A=500(1.015)^t[/tex]  (see what I meant about not having to use the formula with "n" in it if n  1?)

That formula is the answer to part a.  For part b, we are to find how long it takes for the account to reach $800.  $800 goes in for A:

[tex]800=500(1.015)^t[/tex]

Begin by dividing both sides by 500 to get:

[tex]1.6=(1.015)^t[/tex]

The only way to bring that t down from its current exponential position is to take the natural log of both sides.  I will do that and at the same time apply the power rule for logs which says the exponent will come down out in front of the log:

[tex]ln(1.6)=tln(1.015)[/tex]

Divide both sides by ln(1.015):

[tex]\frac{ln(1.6)}{ln(1.015)}=t[/tex]

Do this on your calculator to find that

t = 31.5 years

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