Natural Logs, adding variable exponents.

[Deagonx]

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?

 

[Mark44]

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: e^2x * 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.

a^m * aⁿ = a^{m + n}

What you wrote is e²x + 2. What I think you meant was e^{2x + 2}, which is correct.

If you write this as inline text, you HAVE TO USE PARENTHESES!
I.e., e^(2x + 2)

 

[e^(i Pi)+1=0]

Lol, I think someone is getting missing parentheses fatigue.

 

[micromass]

Lol, I think someone is getting missing parentheses fatigue.

It is really, really, really annoying
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.

 

[solve]

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

 

[Infinitum]

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 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!

 

[solve]

Not exactly butchering. But it is still unclear 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.

 

[Infinitum]

I mean: lne^(2x) * lne^(2). That way I'd get 4x as opposed to e^(2x+2)

See? These are the misunderstanding problems

 

[solve]

See? These are the misunderstanding problems

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.

 

[Mark44]

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 e^2x * e², which is equal to e^{2x + 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?

 

[solve]

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) ?

 

[Mark44]

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.

e^2x * e² = e^{2x + 2}

You can take the ln of both sides to get:
ln(e^2x * e²) = ln(e^{2x + 2})

You cannot rewrite the left side as ln(e^2x) * 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(e^2x) + ln(e²) = 2x + 2

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

 

[solve]

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

e^2x * e² = e^{2x + 2}

You can take the ln of both sides to get:
ln(e^2x * e²) = ln(e^{2x + 2})

You cannot rewrite the left side as ln(e^2x) * 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(e^2x) + 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.

 

[Ray Vickson]

Nice breakdown.

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

Thanks.

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

RGV

 

[solve]

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

RGV

Looks very interesting. Thanks.

 

[Deagonx]

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

 

[Mark44]

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."