# Root of equation with 2 exponentials

1. Mar 5, 2010

### ephedyn

1. The problem statement, all variables and given/known data

This isn't really a homework question, but I was considering a different model of library fines... where the fine at the end of the 1st late day is $0.01, and it increases geometrically with a common ratio of 2, so it goes like$0.01, $0.03,$0.07, $0.15... Then I was comparing this with a typical model of$0.10 additional fines with each day

$0.10,$0.20, $0.30,$0.40...

2. Relevant equations/attempt at a solution

Now, I can work out the day (let n denote this term) when the accumulated fines on the geometric model exceeds the arithmetic model by plotting the graph, where I get

0.1 + 0.1(n-1) < 0.01(2n-1)
10n - 2n + 1 < 0

which has integer solutions for n = 6 or greater. My question is, how do you algebraically solve these sort of equations where the variable is involved twice, as an index and as a base? For example...

x3 - ex = 0

2. Mar 5, 2010

### Brown399

Well, what exactly do you mean when you say "how would you solve for this?" -- ?

If you wanted to find out the number of days which it took for the fine to raise to, say, five dollars (5) you could have to first take the natural log of both sides to get e^x to become x.
and then simply solve

x^3 - e^x = 5

x^3 - x = ln 5

x^3 = x + ln 5

divide by x

x^2 = ln 5

x = sqrt ( ln 5)

but perhaps you mean something els.

3. Mar 5, 2010

### Mentallic

Brown399 you've broken way too many algebraic rules. I suggest you go back and revise that subject.

e.g.
Going from
$$x^3 - e^x = 5$$
to
$$x^3-x=ln(5)$$
is wrong.

From
$$x^3 = x + ln 5$$
to
$$x^2=ln(5)$$
is very wrong.

ephedyn, your problem can't be solved in terms of elementary functions. Basically, you'll only be able to get a good approximation at best.

4. Mar 5, 2010

### Mentallic

You can solve these types of problems in terms of the Lambert W function though.

This says that if $y=xe^x$ then $x=W(y)$.

So, for your example $x^3=e^x$

$$1=x^3e^{-x}$$

$$1=xe^{\frac{-x}{3}}$$

$$\frac{-1}{3}=\frac{-x}{3}e^{\frac{-x}{3}}$$

Hence, $$\frac{-x}{3}=W\left(\frac{-1}{3}\right)$$

$$x=-3W\left(\frac{-1}{3}\right)$$

5. Mar 11, 2010

### ephedyn

^Brown399: ....mmm.

^Mentallic: Ohh! I get how that works now. Thanks a lot. I think there are two complex solutions too but I figured how to work those out.