Math Problem that needs to be solved asap (Pre Calculus)

[Ides_of_March]

My final is in 2 hours and I solved this problem in a lecture in class the other day, but I lost my notes and it was said by my profesor that a question like it will be on the final. It has to do with natural logs.

Find the solution to the problem

The Problem is: 1+3e^-x=7

I've gotten to the point where i've used the properties of logarithms to get to ln2=-x, but -1n-2 doesn't seem to work when I plug it back into the original equation.

My only guess is since the logarithm equals a negative number it's solution is Null. Any confirmation would be apprectiated.

 

[Ace Emerald]

My final is in 2 hours and I solved this problem in a lecture in class the other day, but I lost my notes and it was said by my profesor that a question like it will be on the final. It has to do with natural logs.

Find the solution to the problem

The Problem is: 1+3e^-x=7

I've gotten to the point where i've used the properties of logarithms to get to ln2=-x, but -1n-2 doesn't seem to work when I plug it back into the original equation.

My only guess is since the logarithm equals a negative number it's solution is Null. Any confirmation would be apprectiated.

Why would you change the ln(2) to -ln(-2) it should just be -ln(2) right? Not exactly the best place to ask but my exams are over whatever I'll answer ^_^ you can't log a negative number, and I don't see a reason to make the 2 negative, ln(2) is a real number that you can multiply by negative 1, don't mess with the 2. At least I think that's right precalc was a while ago lol

 

[Codraroll]

http://www.wolframalpha.com/input/?i=1+3e^-x=7

For a further explanation, try some substitution:

1 + 3 e^(-x) = 7.

Let's write u = (-x), to make things simpler.
We can also subtract 1 from each side of the equation.

3e^u = 6

Divide by 3 on both sides of the equation...

e^u = 2

And do the logs...

ln(e^u) = ln(2)

ln(e^a) = a for all a (by definition), so..

u = ln(2)

Now, remember that u = (-x)...

(-x) = ln(2)

Last, multiply (or divide) by -1 to get the answer.

x = -ln(2).

Hope I was of help.

 

[xenu]

subtract 1 from either side;
3e^-x = 6
e^-x = 2
-x = ln 2
x = -ln 2

as for plugging it back in to the original equation,

1 + 3(e^-(-ln 2))
= 1 + 3(e^ln 2)
= 1 + 3(2)
= 1 + 6
= ???

 

[Ides_of_March]

THank you. Love you all. ALso I meant to put that as ln(2) misclick. You're all awesome.