Math help please

[Lager]

What is the solution to this?

e^x+In4 = 3e^x

Thanks for the help. I'm stupid.

 

[bleeb]

We need some clarification:

e^(x+ln4) = 3e^x?

 

[Lager]

Originally posted by: bleeb
We need some clarification:

e^(x+ln4) = 3e^x?

Does it matter? The book doesn't have the parenthesis on it.

Thanks for your help.

 

[Lager]

I #$%&ing hate math.

 

[bleeb]

Originally posted by: coldcut

Originally posted by: bleeb
We need some clarification:

e^(x+ln4) = 3e^x?

Does it matter? The book doesn't have the parenthesis on it.

Thanks for your help.

It could be e^(x) + ln 4 = 3*e^x which is different than e^(x+ln4) = 3*e^x

 

[FrogDog]

Originally posted by: coldcut

Originally posted by: bleeb
We need some clarification:

e^(x+ln4) = 3e^x?

Does it matter? The book doesn't have the parenthesis on it.

Thanks for your help.

We need to know if the '+ln4' part is in the exponent or not.

 

[Lager]

Yes, the x+In4 is part of the exponent.

Sorry about that.

It is e^(x+In4) = 3e^x

 

[GiLtY]

Originally posted by: bleeb

Originally posted by: coldcut

Originally posted by: bleeb
We need some clarification:

e^(x+ln4) = 3e^x?

Does it matter? The book doesn't have the parenthesis on it.

Thanks for your help.

It could be e^(x) + ln 4 = 3*e^x which is different than e^(x+ln4) = 3*e^x

Exactly, depends on whether it's included inside the parenthesis changes the problem quite a bit. If it's the first case then the problem is actually quite easy.

 

[GiLtY]

First break e^(x + ln4) into e ^(x) * e ^(ln4), then it should be easy from there.

 

[Lager]

Originally posted by: GiLtY
First break e^(x + ln4) into e ^(x) * e ^(ln4), then it should be easy from there.

Could you solve it for me? My book doesn't have example and I'm math illiterate.

 

[dopcombo]

erm, if ....

damn. forgot 😛

 

[Lager]

Originally posted by: dopcombo
erm, if

e(x+ln4)=3e(x)
then
e(x+ln4)=e(3x)
then
x+ln4=3x
2x=ln4
x=(ln4)/2
reach for calculator....


damn. where's my casio.

That's it? Thanks for the clarifying that.

 

[GiLtY]

Originally posted by: coldcut

Originally posted by: dopcombo
erm, if

e(x+ln4)=3e(x)
then
e(x+ln4)=e(3x)
then
x+ln4=3x
2x=ln4
x=(ln4)/2
reach for calculator....


damn. where's my casio.

That's it? Thanks for the clarifying that.

I don't think that transition is correct.. you are multiplying 3 by e^x, you can't just put the coefficient into the exponent.

 

[bleeb]

Here is what I get:

e^(x+ln4) = 3e^x

(e^x)(e^ln4) - (3e^x) = 0

4e^x - 3e^x = 0

e^x = 0

but then I get stuck because you can't take the natural log of zero.

 

[Lager]

Originally posted by: bleeb
Here is what I get:

e^(x+ln4) = 3e^x

(e^x)(e^ln4) - (3e^x) = 0

4e^x - 3e^x = 0

e^x = 0

but then I get stuck because you can't take the natural log of zero.

Thanks for the effort. I'll try again and see what I get. Thanks.

 

[GiLtY]

Originally posted by: bleeb
Here is what I get:

e^(x+ln4) = 3e^x

(e^x)(e^ln4) - (3e^x) = 0

4e^x - 3e^x = 0

e^x = 0

but then I get stuck because you can't take the natural log of zero.

Bleeb got it, the answer does not exist (or is not real number).

I don't know what math you are in, there might be answers for it in higher mathematics, but as far as I know (up to Calculus), that's the answer.

Edit: Theoretically the answer should be negative infinity. At negative infinity ln4 can be ignored, and both terms will be zero. But that's probably not the answer you should be looking for. Bleeb's answer is still valid

 

[bleeb]

If its (e^x) + ln 4 = 3e^x

then

ln 4 = 2e^x

e^x = ((ln 4)/2)

x = ln ((ln 4)/2)

x = - 1.8911943935288965945744790181273 (Using WinBlows XP Mis-calculator)

EDIT: To GiLTy. I don't like giving "no solution" as an answer. So perhaps we did this wrong and just need to find a key step.

To ColdCut: Look in your solutions section of your mathematics text book and see if there are "no solution" answers to other problems. I'm assuming this is a lower course mathematics problem (pre-calculus?) and my surmise is that they wouldn't try to trick you like this unless there was a reason to.

 

[GiLtY]

For both Bleep and Coldcut: What math are you guys taking (if you are taking any at all)?

--GiLtY

 

[bleeb]

Originally posted by: GiLtY
For both Bleep and Coldcut: What math are you guys taking (if you are taking any at all)?

--GiLtY

I failed mathematics and am considering becoming a porn star. No need for math in that business. =)

EDIT: YAY, I finally entered into Platinum Status.

 

[GiLtY]

Originally posted by: bleeb

Originally posted by: GiLtY
For both Bleep and Coldcut: What math are you guys taking (if you are taking any at all)?

--GiLtY

I failed mathematics and am considering becoming a porn star. No need for math in that business. =)

EDIT: YAY, I finally entered into Platinum Status.

Hehe, nice.. But you never know, one day you might want to count the # of strokes you take to... then math would come in quite useful 😛

 

[GiLtY]

Originally posted by: bleeb
If its (e^x) + ln 4 = 3e^x

then

ln 4 = 2e^x

e^x = ((ln 4)/2)

x = ln ((ln 4)/2)

x = - 1.8911943935288965945744790181273 (Using WinBlows XP Mis-calculator)

EDIT: To GiLTy. I don't like giving "no solution" as an answer. So perhaps we did this wrong and just need to find a key step.

To ColdCut: Look in your solutions section of your mathematics text book and see if there are "no solution" answers to other problems. I'm assuming this is a lower course mathematics problem (pre-calculus?) and my surmise is that they wouldn't try to trick you like this unless there was a reason to.

"no solution" is a legitimate answer. But strictly speaking negative infinity is the answer.

 

[bleeb]

Originally posted by: GiLtY

Originally posted by: bleeb

Originally posted by: GiLtY
For both Bleep and Coldcut: What math are you guys taking (if you are taking any at all)?

--GiLtY

I failed mathematics and am considering becoming a porn star. No need for math in that business. =)

EDIT: YAY, I finally entered into Platinum Status.

Hehe, nice.. But you never know, one day you might want to count the # of strokes you take to... then math would come in quite useful 😛

hmmm... that might be a good way to increase my stamina and endurance.

 

[vtqanh]

another approach:
  x+ln4        x
e         = 3e
  x+ln4    ln3  x
e         =e  x  e
  x+ln4     x+ln3
e  =        e
x+ln4 = x+ln3
ln4 = ln3
which means that "no solution" is the answer

 

[Lager]

I'm in Trigonometry/College Algebra combined.

I don't want to fail math and become a porn star.

 

[MichaelD]

Originally posted by: GiLtY
First break e^(x + ln4) into e ^(x) * e ^(ln4), then it should be easy from there.

Amazing. The way you said ^^that^^, "should be easy from there."

To me, it's the equivalent of dropping an apple on my desk and saying "map the photonucleoclonic ion structure...in Greek." :Q 😕 :Q to put it mildly.