# Natural Exponential Function Problems

1. Sep 10, 2009

### Gattz

1. The problem statement, all variables and given/known data
10(1 + e-x)-1=3

2. Relevant equations

3. The attempt at a solution

I'm supposed to solve for x, but I don't know how to go about this. I tried dividing the 3 by the 10, but after that I don't know what to do. I believe I should use ln on both side, but that's after I solved for x right?

1. The problem statement, all variables and given/known data
a) 2<lnx<9
b) e2-3x>4

2. Relevant equations

3. The attempt at a solution
It asks me to solve for the inequality of x, but I'm I don't know what that means and I don't know what the greater/less than signs mean.

2. Sep 10, 2009

### Staff: Mentor

Multiply both sides by (1 + e-x), then divide both sides by 3. Don't take the ln of both sides until you have the exponential term all by itself on one side.
For a) it means that you need to arrive at an inequality of the form A < x < B. I really hope you didn't mean you don't understand what these inequality signs mean. If so, you're going to have to go back and review the section that introduced inequalities.

Without working the problem for you, if you had this equation 5 = ln x, you could "exponentiate" each side of the equation; that is, you can make each side the exponent on e, giving you e5 = eln x.
Hopefully, you know that eln x = x, so we have solved this equation for x.

You can do the same thing with your inequality.

For b, you can take the ln of both sides.

3. Sep 10, 2009

### Elucidus

For the second equation, you are expected to exploit the monotonocity of the exponential and logarithmic functions. Specifically:

$u < v \Rightarrow \ln(u) < \ln(v)$

$u < v \Rightarrow e^u < e^v$

--Elucidus

4. Sep 10, 2009

### Staff: Mentor

Nit: "For the second inequality..."

5. Sep 11, 2009

### Elucidus

Point taken. It goes to show that I have way too much math on the brain - I meant to write "second question."

--Elucidus