# Isolating A Variable

1. Sep 11, 2009

### Lancelot59

1. The problem statement, all variables and given/known data
I need to solve a compound interest problem. Easy enough on it's own. However in this case 400 is taken away each month.

So I used this formula:

then subtracted this from it:

(W x 12)t

Where W is the amount withdrawn monthly, and then multiplied by twelve to make it an annual amount.

The initial amount was 10000, the interest rate is 0.06, it gets compounded monthly, and the number of years is the unknown.

So to solve for t I filled in the values:

0 = (10000(1-(0.06-12))^12t) - (400 X 12)t

Then I did this:

48000t/10000 = 1-(0.06-12))^12t

Which went down to:

4.8t = (0.995)^12t

I tried logs and a few other random things, but nothing worked. How can I isolate t?

2. Sep 12, 2009

### slider142

Your equation can be rewritten in the commonly encountered form:
$$-12t\ln (0.995)e^{-12t\ln (0.995)} = -\frac{12}{4.8}\ln(0.995)$$
which simplifies to
$$t = \frac{W\left(-\frac{12}{4.8}\ln(0.995)\right)}{-12\ln(0.995)}$$
where W is the Lambert W function, also known as the product logarithm. The product logarithm is a transcendental function, so it cannot be written as a finite combination of elementary functions. You can ascertain the value of W by looking it up in a table or using a calculator that includes the W function (ie., Mathematica).

3. Sep 12, 2009

### Lancelot59

I didn't get any of that. Is there no simpler way of isolating t in

4.8t = (0.995)^12t

?

4. Sep 12, 2009

### slider142

Unfortunately not. It is in the classic form of a product logarithm (exponential and polynomial in t). Other methods include using normal calculus tools.

5. Sep 12, 2009

### Lancelot59

Ah, so I haven't forgotten everything over the summer.

Could you please explain that method in greater detail?

6. Sep 12, 2009

### Mentallic

I doubt you would be expected to be able to use the W lambert function (whatever the hell that is ).
If it truly came down to isolating such a crazy little nutter, how about an approximation from newton's method or just using a calculator? In the end, this is a money question so an approximation is going to be given anyway.

Heres a nicer way of answering the question though:

let P be the initial payment of \$10,000
and T(n) stand for the nth month of payment

Therefore,
$$T(0)=P$$
$$T(1)=P(0.06)-400$$
$$T(2)=[T(1)](0.06)-400=[P(0.06)-400](0.06)-400$$
$$T(3)=[T(2)](0.06)-400=(\left[P(0.06)-400](0.06)-400\right)-400$$

Expand this out and see if you can find a pattern
You'll have to go back and remember what the formula for the sum of a geometric series is though.
Once you find the pattern (if you do), then try apply use it to express T(t) where t is the number of months since the deposit was made.

Also, even though you forgot to mention that after time t there is nothing left in the bank, you've showed it when using that formula A=0. Similarly, T(t)=0.

Good luck! and if you get stuck don't forget to ask!

7. Sep 12, 2009

### slider142

The Lambert W function is a function with very nice properties. You can really only learn certain values of Lambert by normal study (akin to how one learns natural logarithms or trigonometric functions). The rest, just use your calculator or a series method from calculus. See "A New Elementary Function for our Curricula?" and this MathWorld entry.
Mentallic's method is also a good approach.

8. Sep 12, 2009

### Lancelot59

Ok, I'll look into it and come back. The thing is I wanna kick the calculator as much as possible.

The reason I need to isolate it is so that I can write a program that solves this program accurately.

Newtons' Approximation for this situation sounds like the best plan. Should be easy enough to code it in and make it repeat until it hits zero. So...time to derive it.

Last edited: Sep 12, 2009