Natural Exponential Function Problems

Gattz
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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) e2-3x>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.
 
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Gattz said:

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.
Gattz said:
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) e2-3x>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 = 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.
 
For the second equation, you are expected to exploit the monotonocity of the exponential and logarithmic functions. Specifically:

u &lt; v \Rightarrow \ln(u) &lt; \ln(v)

u &lt; v \Rightarrow e^u &lt; e^v

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

u &lt; v \Rightarrow \ln(u) &lt; \ln(v)

u &lt; v \Rightarrow e^u &lt; e^v

--Elucidus
Nit: "For the second inequality..."
 
Mark44 said:
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." :redface:

--Elucidus
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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