- #1

- 22

- 0

## 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?

- Thread starter Deagonx
- Start date

- #1

- 22

- 0

e^2x * e^2

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?

- #2

Mark44

Mentor

- 33,745

- 5,434

You really need more parentheses if you write this as plain text. I'm assuming that you mean## Homework Statement

e^2x * e^2

e^(2x) * e^2

You can also write exponents (click

Like this: e

But there are theorems that do, I'm sure.## 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?

a

What you wrote is e

If you write this as inline text, you HAVE TO USE PARENTHESES!

I.e., e^(2x + 2)

- #3

- 245

- 0

Lol, I think someone is getting missing parentheses fatigue.

- #4

- 22,097

- 3,283

It is really, really, really annoyingLol, I think someone is getting missing parentheses fatigue.

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

- 94

- 0

What would happen if I did something like this here: lne^2x * lne^2? Would I be butchering math?

- #6

- 881

- 40

Not exactly butchering. But it is still unclear :tongue2: ItWhat would happen if I did something like this here: lne^2x * lne^2? Would I be butchering math?

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

since you have options like X

- #7

- 94

- 0

I mean: lne^(2x) * lne^(2). That way I'd get 4x as opposed to e^(2x+2)Not exactly butchering. But it is still unclear :tongue2: Itcouldmean x*ln(e^2)*ln(e^2)...

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

since you have options like X^{2}in the advanced reply text boxes, andLaTeX, I don't see why you should write such math equations in the first place!

Yeah, I'll check it out.

- #8

- 881

- 40

See? These are the misunderstanding problemsI mean: lne^(2x) * lne^(2). That way I'd get 4x as opposed to e^(2x+2)

- #9

- 94

- 0

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.See? These are the misunderstanding problems

- #10

Mark44

Mentor

- 33,745

- 5,434

I'm not sure what you're asking here. The original problem was eSo,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.

Am I missing what you're asking?

- #11

- 94

- 0

I think I have gone through something like this before. Once I had to simplify: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?

(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:

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

Mark44

Mentor

- 33,745

- 5,434

Yes, it does. Here's where things break down.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) ?

e

You can take the ln of both sides to get:

ln(e

You cannot rewrite the left side as ln(e

There

ln(e

Simplifying the left side gives

2x + 2 = 2x + 2, which is identically true.

- #13

- 94

- 0

Nice breakdown.Yes, it does. Here's where things break down.

e^{2x}* e^{2}= e^{2x + 2}

You can take the ln of both sides to get:

ln(e^{2x}* e^{2}) = ln(e^{2x + 2})

You cannot rewrite the left side as ln(e^{2x}) * ln(e^{2}). There is no property of logs that says that log(AB) = logA * logB

Thereisa property that says that log(AB) = logA + logB, so we can rewrite the equation above as

ln(e^{2x}) + ln(e^{2}) = 2x + 2

Simplifying the left side gives

2x + 2 = 2x + 2, which is identically true.

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

Thanks.

- #14

Ray Vickson

Science Advisor

Homework Helper

Dearly Missed

- 10,706

- 1,728

Note: [tex] X \log(B) = \log(B^X) \text{ if } B > 0.[/tex] Apply this to [itex] X = \log(A)[/itex] to getNice breakdown.

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

Thanks.

[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

- 94

- 0

Looks very interesting. 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

- #16

- 22

- 0

- #17

Mark44

Mentor

- 33,745

- 5,434

I assume that by "variable" you really mean "exponent."

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 2K

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 955

- Last Post

- Replies
- 5

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 6K

- Last Post

- Replies
- 10

- Views
- 2K

- Last Post

- Replies
- 12

- Views
- 8K

- Last Post

- Replies
- 6

- Views
- 1K