Exponential functions/finance problem

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In summary: What interest rate, compounded monthly, is needed in order to double a $10,000 investment in 4 years?In summary, the person needs to compound their interest at a monthly rate of 2.
  • #1
supernova1203
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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 :)
 
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  • #2
To continue with 1.6=(1+i)^20, use logarithms.
One rule: Log[x^y] = y Log[x]
 
  • #3
Atran said:
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.
 
  • #4
This is the formula I use for interest compounded k times a year, which is more or less what you said:
[tex]A = P\left( 1 + \frac{r}{k} \right)^{kt}[/tex]
So,
[tex]16,000 = 10,000\left( 1 + \frac{r}{4} \right)^{4(5)}[/tex]

Is this a homework question? What math class are you in?
 
  • #5
eumyang said:
This is the formula I use for interest compounded k times a year, which is more or less what you said:
[tex]A = P\left( 1 + \frac{r}{k} \right)^{kt}[/tex]
So,
[tex]16,000 = 10,000\left( 1 + \frac{r}{4} \right)^{4(5)}[/tex]

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?
 
  • #6
2x = y <=> log2(y) = x
10x = y <=> log10(y) = x
ex = y <=> loge(y) = ln(y) = x
ax = y <=> loga(y) = x

Log Rules:
Log[a*b] = Log[a] + Log
Log[a/b] = Log[a] - Log
Log[ab] = b * Log[a]
Loga = Logn / Logn[a]
 
  • #7
eumyang said:
This is the formula I use for interest compounded k times a year, which is more or less what you said:
[tex]A = P\left( 1 + \frac{r}{k} \right)^{kt}[/tex]
So,
[tex]16,000 = 10,000\left( 1 + \frac{r}{4} \right)^{4(5)}[/tex]

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

I wonder if you can solve this eumyang @_@
 
  • #8
supernova1203 said:
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" [Broken].
 
Last edited by a moderator:
  • #9
eumyang said:
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" [Broken].

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.
 
Last edited by a moderator:
  • #10
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:
[tex]20,000 = 10,000\left( 1 + \frac{r}{12} \right)^{12(4)}[/tex]

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

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

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

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

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

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

You can finish from here.
 

What is an exponential function?

An exponential function is a mathematical function in which the independent variable is an exponent. The general form of an exponential function is y = ab^x, where a is the initial value and b is the base. It is commonly used to model situations where a quantity grows or decays at a constant percentage rate.

How are exponential functions used in finance?

Exponential functions are frequently used in finance to model compound interest. For example, a bank account that earns 5% interest annually can be modeled using the exponential function y = 100(1.05)^x, where x is the number of years the money is invested. This function allows for easy calculation of the future value of the account.

What is the difference between exponential growth and decay?

Exponential growth occurs when a quantity is increasing at a constant percentage rate, while exponential decay occurs when a quantity is decreasing at a constant percentage rate. In both cases, the rate of change is directly proportional to the current value of the quantity.

How can I determine the growth or decay rate from an exponential function?

The growth or decay rate of an exponential function can be determined by finding the value of the base, b. If b is greater than 1, then the function represents exponential growth, and the growth rate is equal to b-1. If b is between 0 and 1, then the function represents exponential decay, and the decay rate is equal to 1-b.

What are some real-life applications of exponential functions?

Exponential functions have many real-life applications, including population growth, radioactive decay, and the spread of diseases. In finance, they are used to model compound interest, stock market growth, and inflation. They are also used in natural sciences to model growth of microorganisms and chemical reactions.

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