Natural Logs, adding variable exponents.

  • Thread starter Deagonx
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  • #1
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Homework Statement



e^2x * e^2

Homework Equations





The Attempt at a Solution



Really, I'm just a little bit confused on how I would go about adding the exponent. None of the examples in the book cover that. Is it simply e^2x+2?
 

Answers and Replies

  • #2
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Homework Statement



e^2x * e^2
You really need more parentheses if you write this as plain text. I'm assuming that you mean
e^(2x) * e^2

You can also write exponents (click Go Advanced to open the advanced menu, and then click the X2 button).

Like this: e2x * e2

Homework Equations





The Attempt at a Solution



Really, I'm just a little bit confused on how I would go about adding the exponent. None of the examples in the book cover that. Is it simply e^2x+2?
But there are theorems that do, I'm sure.

am * an = am + n

What you wrote is e2x + 2. What I think you meant was e2x + 2, which is correct.

If you write this as inline text, you HAVE TO USE PARENTHESES!
I.e., e^(2x + 2)
 
  • #3
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Lol, I think someone is getting missing parentheses fatigue.
 
  • #4
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Lol, I think someone is getting missing parentheses fatigue.
It is really, really, really annoying :biggrin:
I think you can easily distinguish a person who knows math from a person who doesn't know math by seeing how they handle parantheses.
 
  • #5
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What would happen if I did something like this here: lne^2x * lne^2? Would I be butchering math?
 
  • #6
881
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What would happen if I did something like this here: lne^2x * lne^2? Would I be butchering math?
Not exactly butchering. But it is still unclear :tongue2: It could mean x*ln(e^2)*ln(e^2)...

ln(e^(2x))*ln(e^2) is more appropriate, I think.

since you have options like X2 in the advanced reply text boxes, and LaTeX, I don't see why you should write such math equations in the first place!
 
  • #7
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Not exactly butchering. But it is still unclear :tongue2: It could mean x*ln(e^2)*ln(e^2)...

ln(e^(2x))*ln(e^2) is more appropriate, I think.

since you have options like X2 in the advanced reply text boxes, and LaTeX, I don't see why you should write such math equations in the first place!
I mean: lne^(2x) * lne^(2). That way I'd get 4x as opposed to e^(2x+2)

Yeah, I'll check it out.
 
  • #8
881
40
I mean: lne^(2x) * lne^(2). That way I'd get 4x as opposed to e^(2x+2)
See? These are the misunderstanding problems :wink:
 
  • #9
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See? These are the misunderstanding problems :wink:
So,is lne^(2x) * lne^(2)= 4x acceptable just like e^(2x+2) is? Because, I am thinking I could do lne^(2x+2)=2x+2 and don't see how 2x+2 equals 4x.
 
  • #10
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So,is lne^(2x) * lne^(2)= 4x acceptable just like e^(2x+2) is? Because, I am thinking I could do lne^(2x+2)=2x+2 and don't see how 2x+2 equals 4x.
I'm not sure what you're asking here. The original problem was e2x * e2, which is equal to e2x + 2, and that's not much more you can do with it. You can't just come in an take the log of each factor in a product. Since 2x + 2 is not identically equal to 4x, it should be clear that ln(e^(2x)) * ln(e^2) ≠ ln(e^(2x + 2)).

Am I missing what you're asking?
 
  • #11
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You can't just come in an take the log of each factor in a product. Since 2x + 2 is not identically equal to 4x, it should be clear that ln(e^(2x)) * ln(e^2) ≠ ln(e^(2x + 2)).

Am I missing what you're asking?
I think I have gone through something like this before. Once I had to simplify:

(x^2-y^2)^(1/2)* (x-y)^(3/2)* (x+y)^(-1/2).

What I did was:

(x^2-y^2)^(1/2*2)* (x-y)^(3/2*2)* (x+y)^(-1/2*2)

I just couldn't understand why my "simplification" didn't produce the right answer till I re-framed what I did in this manner:

[(x^2-y^2)^(1/2)* (x-y)^(3/2)* (x+y)^(-1/2)]^2

Only then I could see I was just squaring the whole expression. It was like squaring x^(2) and not quiet getting why x^(2) doesn't equal x^(4).

Does what I wrote above sound something like me coming and taking the log of each factor in a product in e^(2x)* e^(2) ?
 
  • #12
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5,434
I think I have gone through something like this before. Once I had to simplify:

(x^2-y^2)^(1/2)* (x-y)^(3/2)* (x+y)^(-1/2).

What I did was:

(x^2-y^2)^(1/2*2)* (x-y)^(3/2*2)* (x+y)^(-1/2*2)

I just couldn't understand why my "simplification" didn't produce the right answer till I re-framed what I did in this manner:

[(x^2-y^2)^(1/2)* (x-y)^(3/2)* (x+y)^(-1/2)]^2

Only then I could see I was just squaring the whole expression. It was like squaring x^(2) and not quiet getting why x^(2) doesn't equal x^(4).

Does what I wrote above sound something like me coming and taking the log of each factor in a product in e^(2x)* e^(2) ?
Yes, it does. Here's where things break down.

e2x * e2 = e2x + 2

You can take the ln of both sides to get:
ln(e2x * e2) = ln(e2x + 2)

You cannot rewrite the left side as ln(e2x) * ln(e2). There is no property of logs that says that log(AB) = logA * logB

There is a property that says that log(AB) = logA + logB, so we can rewrite the equation above as
ln(e2x) + ln(e2) = 2x + 2

Simplifying the left side gives
2x + 2 = 2x + 2, which is identically true.
 
  • #13
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Yes, it does. Here's where things break down.

e2x * e2 = e2x + 2

You can take the ln of both sides to get:
ln(e2x * e2) = ln(e2x + 2)

You cannot rewrite the left side as ln(e2x) * ln(e2). There is no property of logs that says that log(AB) = logA * logB

There is a property that says that log(AB) = logA + logB, so we can rewrite the equation above as
ln(e2x) + ln(e2) = 2x + 2

Simplifying the left side gives
2x + 2 = 2x + 2, which is identically true.
Nice breakdown.

My brain has a tendency to see equations where there aren't any. It's learning, though.

Thanks.
 
  • #14
Ray Vickson
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Nice breakdown.

My brain has a tendency to see equations where there aren't any. It's learning, though.

Thanks.
Note: [tex] X \log(B) = \log(B^X) \text{ if } B > 0.[/tex] Apply this to [itex] X = \log(A)[/itex] to get
[tex] \log(A) \cdot \log(B) = \log(B^{\log(A)}).[/tex] This also equals
[tex] \log(A^{\log(B)}),[/tex]
so, apparently we have
[tex] A^{\log(B)} = B^{\log(A)}[/tex] for all A, B > 0. Hmmm...I've never seen that before, but it does check out.

RGV
 
  • #15
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Note: [tex] X \log(B) = \log(B^X) \text{ if } B > 0.[/tex] Apply this to [itex] X = \log(A)[/itex] to get
[tex] \log(A) \cdot \log(B) = \log(B^{\log(A)}).[/tex] This also equals
[tex] \log(A^{\log(B)}),[/tex]
so, apparently we have
[tex] A^{\log(B)} = B^{\log(A)}[/tex] for all A, B > 0. Hmmm...I've never seen that before, but it does check out.

RGV
Looks very interesting. Thanks.
 
  • #16
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Sorry guys. I'm really not used to writing equations in text format. If I wanted to show that the 2x+2 was a variable on paper I would just make it higher and small. ;P
 
  • #17
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Sorry guys. I'm really not used to writing equations in text format. If I wanted to show that the 2x+2 was a variable on paper I would just make it higher and small. ;P
I assume that by "variable" you really mean "exponent."
 

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