Exponential functions/finance problem

[supernova1203]

Homework Statement

Annoying problem, I recently got a graphing calculator (t1-89) and according to instructions i can calculate the i (interest) on it but each problem before i attempt on a graphing calculator i attempt to solve myself manually without a graphing calculator as a safeguard to prevent myself from being handicapped and using the graphing calculator as a crutch but i cannot solve this problem manually @_@

Someone has invested 10000$ for 5 years. At the end of 5 years they would like to have 16000$ to buy a new motorcycle. What rate of interest, compounded quarterly does this person need to achieve this goal?

Homework Equations

A = P(1+i)^n where a is amount, p is principle, i is rate of interest and n is the time(in years usually)

A=P+I (here the I is interest, NOT the rate of interest which is obviously different from interest)

The Attempt at a Solution

A=P(1+i)ⁿ

16000=10000(1+i)⁵ (because of 5 years, but since interest is compounded quarterly n=5X4
n=20)
Similarly we shall need to divide the rate of interest as well by 4 times, that is if i can find it. That is after we divide it by a 100 to get the actual rate itself

16000/10000=(1+i)²⁰
1.6=(1+i)^20
At this point i get stuck, any help would be greatly appreciated :)

 

[Atran]

To continue with 1.6=(1+i)^20, use logarithms.
One rule: Log[x^y] = y Log[x]

 

[supernova1203]

To continue with 1.6=(1+i)^20, use logarithms.
One rule: Log[x^y] = y Log[x]

I have heard of logarithms but have no idea what they are.

 

[eumyang]

This is the formula I use for interest compounded k times a year, which is more or less what you said:
A = P\left( 1 + \frac{r}{k} \right)^{kt}
So,
16,000 = 10,000\left( 1 + \frac{r}{4} \right)^{4(5)}

Is this a homework question? What math class are you in?

 

[supernova1203]

This is the formula I use for interest compounded k times a year, which is more or less what you said:
A = P\left( 1 + \frac{r}{k} \right)^{kt}
So,
16,000 = 10,000\left( 1 + \frac{r}{4} \right)^{4(5)}

Is this a homework question? What math class are you in?

This is a homework question, I am in the 11th grade math class, i got 1 more class (advanced function) i need to take after this course and then i take calculus :) I got all sorts of weird answers doing it manually...can you show me how to get the solution?

 

[Atran]

2^x = y <=> log₂(y) = x
10^x = y <=> log₁₀(y) = x
e^x = y <=> log_e(y) = ln(y) = x
a^x = y <=> log_a(y) = x

Log Rules:
Log[a*b] = Log[a] + Log
Log[a/b] = Log[a] - Log
Log[a^b] = b * Log[a]
Log_a = Log_n / Log_n[a]

 

[supernova1203]

This is the formula I use for interest compounded k times a year, which is more or less what you said:
A = P\left( 1 + \frac{r}{k} \right)^{kt}
So,
16,000 = 10,000\left( 1 + \frac{r}{4} \right)^{4(5)}

Is this a homework question? What math class are you in?

I wonder if you can solve this eumyang @_@

 

[eumyang]

I wonder if you can solve this eumyang @_@

I'm not allowed to -- forum rules. You're going to have to read up on logarithms and their properties if you want to solve this algebraically.

To do this graphically on the TI-89, you'll need to enter the following in the Y= editor:
Y1 = 1.6 and
Y2 = (1 + x/4)^20
Then use the intersection tool. See http://www.prenhall.com/divisions/esm/app/graphing/ti89/graphing/working_with_graphs/F5/F5_Math.html" .

 

[supernova1203]

I'm not allowed to -- forum rules. You're going to have to read up on logarithms and their properties if you want to solve this algebraically.

To do this graphically on the TI-89, you'll need to enter the following in the Y= editor:
Y1 = 1.6 and
Y2 = (1 + x/4)^20
Then use the intersection tool. See http://www.prenhall.com/divisions/esm/app/graphing/ti89/graphing/working_with_graphs/F5/F5_Math.html" .

ugh... i know how to do it on a graphing calculator, but i want to learn how to do it manually that was the whole point...i tried to solve it on my own manually but i couldn't tried like a dozen times...then i posted here...and in my course we haven't studied logarithms yet but i suppose they expect me to solve this through what they call trial and error which i have no idea how that applies to this situation but i suppose i can try to learn about logarithms.

 

[eumyang]

I'll show you how to do a similar problem, then. (I hope I'm not breaking any rules here.)

What interest rate, compounded monthly, is needed in order to double a $10,000 investment in 4 years?
Setup:
20,000 = 10,000\left( 1 + \frac{r}{12} \right)^{12(4)}

Divide by 10,000:
2 = \left( 1 + \frac{r}{12} \right)^{48}

Take the logarithm of both sides. (Base doesn't matter, but I'll use base-10.)
\log 2 = \log \left( 1 + \frac{r}{12} \right)^{48}

Use the property that Atran mentioned (Log[a^b] = b * Log[a]):
\log 2 = 48 \log \left( 1 + \frac{r}{12} \right)

Divide by 48:
\frac{\log 2}{48} = \log \left( 1 + \frac{r}{12} \right)

Use the property if u = v, then b^u = b^v. Since I used base-10 logarithms I'll use 10 as the base:
10^{(\log 2)/48} = 10^{\log \left( 1 + r/12 \right)}

Use the logarithm property b^{\log_b y} = y:
10^{(\log 2)/48} = 1 + \frac{r}{12}

You can finish from here.