General

Thank you sooooo much! I appreciate it!

Sorry was doing homework. Here's the work, I tried to make it simple and easy to understand.

17000 = 9000 ( 1 + i/4)^(2*4)
17000 = 9000 ( 1 + i/4)^(8)
1.8889 = ( 1 + i/4)^(8)
log(1.8889) = log(( 1 + i/4)^(8))
.2762 = 8*log( 1 + i/4)
.0345 = log( 1 + i/4)
log( 1 + i/4) = .0345

This is where you use a log rule. I'll state it if you don't already know it.
log(a) = b can be rewritten as a = 10^b

( 1 + i/4) = 10^.0345
1 + i/4 = 1.0827
i/4 = .0827
i = .331

What grade are you in? Because if you're in like 9th grade, it seems a little too difficult to use logs.
Regardless, I'll explain them.
Taking the log of both sides in an equation is the same as multiplying both sides of the equation by a number.
You take the log of both sides of an equation when you have a variable exponent. Or when an expression with a variable is raised to a power.
For example, it'd be useful to use logs in these equations:
447 = 8^x
447 = (x + 1)^8

When you take the log of something, the exponent becomes the coefficient of the whole log itself.
447 = 8^x
log(447) = log(8^x)
log(447) = xlog(8)

447 = (x + 1)^8
log(447) = log((x + 1)^8)
log(447) = 8log(x + 1)

Ask me anytime when you have more questions or problems.

^ Not really since they have different methods of doing this.

Aw my message didn't go through. "Yes please!"

[quote=DaTruStrDit]<---- rate always kills me since I always do it wrong apparently. A & P is fine but time and rates are just... Derp.[/quote]
a=p(1+i)^n
That's just the general equation. It changes a little depending on how often interest is compounded.
The specific formula is a=p(1+i/x)^(n*x)
where x=the number of times it's compounded in one year.
So if it's annual, x=1, quarterly x=2, monthly x=12, and so on.

17000 = 9000 ( 1 + i/4)^(2*4)
Do you want me to do it for you and show the steps?

<---- rate always kills me since I always do it wrong apparently. A & P is fine but time and rates are just... Derp.

find rate = solve for i. you have a, p and n. try to do the rest yourself.

(b) Now N is 2 years and the question mark is on the rate. How does one find rate? One last time and then I got it!

[quote=DaTruStrDit]Foundations. Not sure what that is for you guys. Also I divided both by 0.1513 and got 1.8225 [/quote]
Sorry it was a typo when I transferred my work to my post. The answer is still right, It's just supposed to be .0513 and not .1513. I missed the '1' key.

Foundations. Not sure what that is for you guys. Also I divided both by 0.1513 and got 1.8225

[quote=DaTruStrDit]^ Close the back of the textbook says 5.5 years. If you round it then it would be 5.4 years. Or should I just accept your answer and erase the back of the book for being wrong? xD[/quote]
5.4 years is a lot more accurate than 5.5 years.
Most teachers would accept both, to be honest.
What math are you in at the moment? Is this for your regular math class, or is this a financial class?

Just simply plug everything into the equation, and isolate "n".

^ Close the back of the textbook says 5.5 years. If you round it then it would be 5.4 years. Or should I just accept your answer and erase the back of the book for being wrong? xD

[quote=DaTruStrDit]Then what the heck do I do with them? Soooo combine 9000 with the right side and log the 17000... Now what? Divide them both?[/quote]
Did my explanation help? You're basically solving for a single variable.

Then what the heck do I do with them? Soooo combine 9000 with the right side and log the 17000... Now what? Divide them both?

[quote=DaTruStrDit]^ Yup did that but not sure what to do with the A and P. Do I subtract or divide to get rid of P first?[/quote]

When you have something like x = a*y -> x/a = y. It's not x - a = y.

^ Yup did that but not sure what to do with the A and P. Do I subtract or divide to get rid of P first?

take log of both sides:

log 17000 = 9000 * n * log (1.12), solve for n

This is the modified interest formula for quarterly compounded interest rates:
A = P(1 + i/4)^(n*4)
A = principal + interest earned
P = principal/initial amount deposited
i = interest rate (decimal)
n = years invested

17000 = 9000 ( 1 + .12/4 ) ^ (4n)
1.8889 = ( 1 + .12/4 ) ^ (4n)
1.8889 = ( 1 + .03 ) ^ (4n)
1.8889 = ( 1.03 ) ^ (4n)
1.8889 = 1.1255 ^ n
log(1.8889) = log(1.1255 ^ n)
.2762 = n*log(1.1255)
.2762 = .0513n
n = 5.379
Not sure if this is right.

^ Nope my teacher basically gave us the homework and said "Learn it yourself!"

do you know how to log? its a formula from logarithm