PLZ HELP ALGEBRA!!!!
AND PLZ SHOW WORK

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