Solving 3e^(x+2) = e^(-x) for x

[CalculusSandwich]

I have the problem 3e^(x+2) = e^(-x)

and I need to find X. It's been a while since I've looked at any Calculus 2 stuff so I am not sure on what to do here.

I take the natural log of both sides

LN(3e^(x+2) =LN(e^(-x))= (x+2)ln3e = -xlne

So (x+2)Ln3e= -x

Where do I go from here?

 

[cristo]

It's easier to collect the terms before you log both sides. Try collecting the exponentials on one side. Do you know, for example, how to simplify $$e^ae^b$$ ?

 

[CalculusSandwich]

It's easier to collect the terms before you log both sides. Try collecting the exponentials on one side. Do you know, for example, how to simplify $$e^ae^b$$ ?

the same was as you would any number x^3 * x^2 = X^5

So e^(x+2) * e^(-x) = (x+2-x) or e^2, correct?

 

[cristo]

yes, $$e^{x+2}e^{-x}=e^2$$ However, when manipulating the above equation, you need to multiply both sides by $$e^{x}$$ in order to cancel the $$e^{-x}$$ on the RHS

 

[Mute]

Also note that ln(3*e^a) is not a*ln(3e), so unless you just didn't put some brackets around (3e)^(x+2), your current expression on the left-hand-side of your last line is incorrect.

 

[HallsofIvy]

I have the problem 3e^(x+2) = e^(-x)

and I need to find X. It's been a while since I've looked at any Calculus 2 stuff so I am not sure on what to do here.

I take the natural log of both sides

LN(3e^(x+2) =LN(e^(-x))= (x+2)ln3e = -xlne

So (x+2)Ln3e= -x

Where do I go from here?

You don't go anywhere. Your first step was wrong: ln(3e^x+2)= ln 3+ (x+2) not (x+2)ln(3e).

Your final equation is ln(3)+ x+2= -x. Would you know how to solve
A+ x+ B= -x for constants A and B?

 

[dextercioby]

I have the problem 3e^(x+2) = e^(-x)

and I need to find X. It's been a while since I've looked at any Calculus 2 stuff so I am not sure on what to do here.

I take the natural log of both sides

LN(3e^(x+2) =LN(e^(-x))= (x+2)ln3e = -xlne

So (x+2)Ln3e= -x

Where do I go from here?

Others pointed out flaws, i just want to be sure that you can solve

$$e^{-2x}=3e^{2}$$

Daniel.

P.S. I get $x=-\frac{1}{2}\ln 3 -1 + ik\pi \ , \ k\in\mathbb{Z}$