- #1
asdf1
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if h=ln(absolute x)
then how do you calculate e^h?
then how do you calculate e^h?
VietDao29 said:It must be |x|. Have you ever seen:
[tex]e ^ {\mbox{something}} = \mbox{a negative number}[/tex]??
Viet Dao,
lurflurf said:yes
[tex]e^{\pi i}=-1[/tex]
What were the preceding steps? This type of thing happens a lot when what you were finding was the absolute value of something and droping it gives the something.asdf1 said:that's what i originally thought,
but the correct answer doesn't need absolute value...
i think that's strange...
Well, you can always take a numerical example if you're in doubt...:asdf1 said:that's what i originally thought,
but the correct answer doesn't need absolute value...
i think that's strange...
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]
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:My example wasn't necessarily over the reals, I don't see why we'd have to exclude to complex numbers - at least not for this example which is only intended to show that abs values are needed.
So it was due to integrating a logarimic derivative.asdf1 said:Originally, I was trying to solve this problem:
dy/dx + ytanx = sinx, y(0)=1
The first step I did was to integrate tanx and then there came the absolute value dilema...
TD said:My example wasn't necessarily over the reals, I don't see why we'd have to exclude to complex numbers - at least not for this example which is only intended to show that abs values are needed.
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.
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~
The value of e^h if h = ln(absolute x) is equal to the absolute value of x. This is because the natural logarithm of a number is the power to which e must be raised to equal that number. So if h = ln(absolute x), then e^h = x, and taking the absolute value of x ensures that the result is always positive.
To calculate e^h if h = ln(absolute x), you can simply raise e to the power of ln(absolute x). This can be written as e^(ln| x|). Since the natural logarithm and exponential functions are inverse of each other, e^(ln| x|) simplifies to |x|.
No, e^h cannot be negative if h = ln(absolute x). This is because the natural logarithm of a positive number is always positive, and raising e to a positive power will always result in a positive number. Moreover, taking the absolute value of x ensures that the result is always positive.
The relationship between e^h and ln(absolute x) is that they are inverse functions of each other. This means that e^h = x if and only if h = ln(absolute x). This relationship is based on the fact that the natural logarithm of a number is the power to which e must be raised to equal that number.
Changing the value of x will directly affect the value of e^h if h = ln(absolute x). This is because the value of e^h is equal to the absolute value of x. This means that as the value of x increases, the value of e^h also increases. Similarly, as the value of x decreases, the value of e^h decreases.