# Thread: e, ln and absolute values

1. ## e, ln and absolute values

Hello,

An example in a text book I am studying uses the line:

$e^{\ln|sec(x)|}=sec(x)$

But I can't see how it could be anything other than:

$e^{\ln|sec(x)|}={|sec(x)|}$

Because $sec(x)$ can sometimes be a negative value and I can't see how $e$ to any exponent could be negative.

Is my way of thinking correct or am I missing something?

2. Originally Posted by sael
Hello,

An example in a text book I am studying uses the line:
$e^{\ln|sec(x)|}=sec(x)$
But I can't see how it could be anything other than:
$e^{\ln|sec(x)|}={|sec(x)|}$
Because $sec(x)$ can sometimes be a negative value and I can't see how $e$ to any exponent could be negative.

Is my way of thinking correct or am I missing something?

Is there any given interval for x in the example?

3. Sorry, still getting used to the forum interface.

4. There is no interval given in the example.

5. Isn't it illegal to take a natural log of a negative value? Could that be why they use the absolute value?

6. Originally Posted by satis
Isn't it illegal to take a natural log of a negative value? Could that be why they use the absolute value?
Yes ln(x) is only for x> 0. My problem is not with that but that the absolute value is not preserved in the text book example and I think it should be :-)

7. in your defense, against the textbook

$e^{\ln|sec(PI)|}=1$
$sec(PI)=-1$

Perhaps you should post more of the example

8. Originally Posted by sael
Hello,

An example in a text book I am studying uses the line:

$e^{\ln|sec(x)|}=sec(x)$

But I can't see how it could be anything other than:

$e^{\ln|sec(x)|}={|sec(x)|}$

Because $sec(x)$ can sometimes be a negative value and I can't see how $e$ to any exponent could be negative.

Is my way of thinking correct or am I missing something?

You are correct, but I suspect that there is a restriction on the domain in the example.

9. Originally Posted by billa
in your defense, against the textbook

$e^{\ln|sec(PI)|}=1$
$sec(PI)=-1$

Perhaps you should post more of the example
The example doesn't belong in this forum but here it is anyway

I get the result $y(x)=c|\cos(x)|- 2\cos^2(x)$ which is discontinuous and their result isn't which makes me wonder what I am missing.

10. Forgetting the example I posted. I am starting to see what I might be missing.

We don't write:
$(x^2)^\frac{1}{2}=\pm{x}$

$(x^2)^\frac{1}{2}=x$

So somehow I have to make the connection to $e^{\ln|\sec{(x)}|}=\sec{(x)}$

11. I recommend you re-post this in the differential equations section... I am currently in a DE and I want to know the answer now... if you get please post it here lol

12. Originally Posted by billa
I recommend you re-post this in the differential equations section... I am currently in a DE and I want to know the answer now... if you get please post it here lol
From what I understand, at the moment, the problem isn't related to the particular ODE. I think it has more to do with $\ln{|x|}$ being defined for all real x where as $e^{x}$ is the inverse for $\ln{x}$ for all $x>0$. So somehow I have to get my mind around functions that are not one-to-one. This is new ground for me and it doesn't quite make sense yet.

13. The best I can come up with is that $e^x$ is not just a number with an exponent but is defined (in my texts at least) in terms of being the inverse of $\ln{x}$ that just happens to behave as though it obeys the laws of exponents from algebra.

The example appears to be treating $e^{\ln|\sec{(x)}|}$ as though it means "inverse of" $\ln{|\sec{(x)}|}$, so in general what I really need is an understanding of the inverse of $\ln{|x|}$ which is not a one-to-one function of x.

My brain hurts

14. Okay this really doesn't belong in this forum but I thought I had better relay what I have found anyway.

It appears that within the example problem the part $e^h=\sec(x)$ assumes that the constant of integration from $h=\int\!p\,dx$ is ignored while evaluating (4). This comes from the derivation of (4) which I shouldn't post here because it is ODE related.

Adding in the constant of integration:

$h=\int\!p\,dx=\int\!\tan (x)\,dx=\ln{|\sec (x)| + c_1}$

$e^h=e^{\ln{|\sec (x)|}+c_1}=e^{c_1}|\sec(x)|$
$e^{c_1}|\sec(x)|=c_2\sec(x)\quad where\,(c_2=\pm{e^{c_1}})$

The constant $c_2$ is ignored while evaluating (4) and hence the absolute value vanishes