Natural Exponential Function Problems

[Gattz]

Homework Statement

10(1 + e^-x)^-1=3

Homework Equations

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?

Homework Statement

a) 2<lnx<9
b) e^2-3x>4

Homework Equations

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.

 

[Mark44]

Homework Statement

10(1 + e^-x)^-1=3

Homework Equations

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.

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.

I believe I should use ln on both side, but that's after I solved for x right?

Homework Statement

a) 2<lnx<9
b) e^2-3x>4

Homework Equations

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.

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 = e^{ln x}.
Hopefully, you know that e^{ln 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.

 

[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

 

[Mark44]

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

Nit: "For the second inequality..."

 

[Elucidus]

Nit: "For the second inequality..."

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

--Elucidus