- #1

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if h=ln(absolute x)

then how do you calculate e^h?

then how do you calculate e^h?

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- #1

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if h=ln(absolute x)

then how do you calculate e^h?

then how do you calculate e^h?

- #2

TD

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Seems simple, do you mean this?

- #3

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but the correct answer doesn't need absolute value...

i think that's strange...

- #4

VietDao29

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[tex]e ^ {\mbox{something}} = \mbox{a negative number}[/tex]??

Viet Dao,

- #5

lurflurf

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yesVietDao29 said:

[tex]e ^ {\mbox{something}} = \mbox{a negative number}[/tex]??

Viet Dao,

[tex]e^{\pi i}=-1[/tex]

- #6

rachmaninoff

- #7

VietDao29

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lurflurf said:yes

[tex]e^{\pi i}=-1[/tex]

Okay, my something is [tex]\in \mathbb{R}[/tex], not C .

Viet Dao,

- #8

lurflurf

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What were the preceding steps? This type of thing happens alot when what you were finding was the absolute value of something and droping it gives the something.asdf1 said:

but the correct answer doesn't need absolute value...

i think that's strange...

for instance if

xy'=y y(0)=0 (the solution being y=x or y=-x)

one might get

|y|=|x| as an answer but though error have

y=|x|

- #9

TD

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Well, you can always take a numerical example if you're in doubt...:asdf1 said:

but the correct answer doesn't need absolute value...

i think that's strange...

[tex]\begin{array}{l}

e^{\ln \left( { - 1} \right)} = - 1 \\

e^{\ln \left| { - 1} \right|} = 1 \\

\end{array}[/tex]

- #10

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This is incorrect. Assuming we're working in the reals [itex]e^{\ln(-1)}[/itex] isn't defined, because [itex]\ln[/itex] is undefined for negative numbers and so any expression including it is also undefined.TD said:Well, you can always take a numerical example if you're in doubt...:

[tex]\begin{array}{l}

e^{\ln \left( { - 1} \right)} = - 1 \\

e^{\ln \left| { - 1} \right|} = 1 \\

\end{array}[/tex]

- #11

TD

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- #12

lurflurf

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Again where how did this arise. One way absolute values often arise in arguments of logarithums is when one wants an antiderivative of a logarothmic derivative that is restricted to reals and valid almost every where.TD said:

ie

xy'=1 y(e)=1

y=log(|x|)

clearly if we are working with complex numbers

y=log(x) would do

and sometimes errors arise as when

xy'=y y(1)=1

clearly

y=x

when restricted to real one often finds

|y|=exp(log(|x|))=|x|

but a common error is to get

y=|x| which satisfys the differential equation every where except the origin

if you log(|x|) did not arise though integrating a logarithmic derivative perhaps some other error was made to cause you to get |h| when h was wanted (or mayber h>0 so it does not matter).

- #13

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dy/dx + ytanx = sinx, y(0)=1

The first step I did was to integrate tanx and then there came the absolute value dilema...

- #14

lurflurf

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So it was due to integrating a logarimic derivative.asdf1 said:

dy/dx + ytanx = sinx, y(0)=1

The first step I did was to integrate tanx and then there came the absolute value dilema...

Say you were to solve th associated homogeneous problem in preparation for variation of parameters.

y'+y tan(x)=0

y'/y=cos'(x)/cos(x)

log(|y|)=a+log(|cos(x)|)

|y|=|C||cos(x)| then let C absorb the sign

y=C*cos(x)

which is clearly a solution

of course if we were working with complex numbers we would never have introduced the absolute value. This problem arises though finding the function. It has nothing to do with the function itself.

- #15

HallsofIvy

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Because the original posterTD said:

- #16

TD

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- #17

VietDao29

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[tex]\ln|x| \in \mathbb{R} \mbox{, \forall x \in \mathbb{R} - \{0\}}[/tex]?????

Viet Dao,

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- #19

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that question seems to work, but in this question:lurflurf said:So it was due to integrating a logarimic derivative.

Say you were to solve th associated homogeneous problem in preparation for variation of parameters.

y'+y tan(x)=0

y'/y=cos'(x)/cos(x)

log(|y|)=a+log(|cos(x)|)

|y|=|C||cos(x)| then let C absorb the sign

y=C*cos(x)

which is clearly a solution

of course if we were working with complex numbers we would never have introduced the absolute value. This problem arises though finding the function. It has nothing to do with the function itself.

y`= -y/x, y(1)=1

the same problem arises, but the constant doesn't absorb the negative sign...

it's weird~

- #20

lurflurf

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What do you meanasdf1 said:that question seems to work, but in this question:

y`= -y/x, y(1)=1

the same problem arises, but the constant doesn't absorb the negative sign...

it's weird~

y'=-y/x y(1)=1

y'/y=-1/x

log(|y|)=-log(|x|)+a=log(1/|x|)+a

|y|=exp(a)/|x|

|y|=|c|/|x|

y=c/x

y(1)=c/1=c=1

y=1/x

- #21

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sorry~

i made a mistake in the calculations...

- #22

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thanks for correcting my mistake~

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