Simplifying Transcendental Functions

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In summary, the expression 13ln[exp[(x^2+2)] - exp[ln(13x^2+26)]] can be reduced to 13(x^2+2)- (13x^2+26) by recognizing that exp[ln(13x^2+26)] is equivalent to 13x^2+26 due to the inverse relationship between exp and ln. Additionally, the function x^(e^x) can be derived using logarithmic differentiation, where the natural log of x^(e^x) is equal to (e^x)x. By using the product rule and solving for y', the derivative can be found.
  • #1
MercuryRising
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Im stuck on these probelms
Simplify
13ln[exp[(x^2+2)] - exp[ln(13x^2+26)]]

exp[ln(13x^2+26) cancels out to 13x^2+26 but i don't see how that helps with the entire probelm
Derive
x^(e^x) my friend suggested power rule...but that seems a litle too simple and would create a quite a mess

thank you in advance
 
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  • #2
For the simplification question, whatever's in the exp() or ln() when one is inside the other gets out because ln and exp are inverses... so you're left to simplify some algebraic expressions! (Some cancelling out happens)

And for finding the derivative of x^(e^x): (assuming it's defined as a function f)
f = x^(e^x)
Take the natural log of both sides
ln f = (e^x) ln x

And use the chain rule and product rule and solve for f'. :)
 
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  • #3
MercuryRising said:
Im stuck on these probelms
Simplify
13ln[exp[(x^2+2)] - exp[ln(13x^2+26)]]

exp[ln(13x^2+26) cancels out to 13x^2+26 but i don't see how that helps with the entire probelm
I presume that if you know that, then you also know that exp[ln(13x^2+ 26)]= 13x^2+ 26 (exp and ln are "inverse" functions and "cancel" each other.

And you don't see how reducing 13ln[exp[(x^2+2)] - exp[ln(13x^2+26)] to 13(x^2+ 2)- (13x^2+ 26) helps at all?? Believe, me that simplifies trivially!



Derive
x^(e^x) my friend suggested power rule...but that seems a litle too simple and would create a quite a mess

thank you in advance

Use "logarithmic differentiation". If y= x^(e^x), then ln y= (e^x)x. On the left, (ln y)'= (1/y)y' and you should be able to differentiate (e^x)x using the product rule. Solve the resulting equation for y'.
 
  • #4
The way it's written reduces to 13ln(e^expression - 13*expression), where expression is x^2+2, so it doesn't reduce trivially. Unless it's written wrong in the OP.
 
  • #5
HallsofIvy said:
I presume that if you know that, then you also know that exp[ln(13x^2+ 26)]= 13x^2+ 26 (exp and ln are "inverse" functions and "cancel" each other.

And you don't see how reducing 13ln[exp[(x^2+2)] - exp[ln(13x^2+26)] to 13(x^2+ 2)- (13x^2+ 26) helps at all?? Believe, me that simplifies trivially!

hmm how exactly do you get to the algebraic expression 13(x^2+ 2)- (13x^2+ 26) the function dave posted was the farthest i got..:confused:
 
  • #6
daveb said:
The way it's written reduces to 13ln(e^expression - 13*expression), where expression is x^2+2, so it doesn't reduce trivially. Unless it's written wrong in the OP.
I think you're misreading the OP's post. Let's write this out in LaTex so there's no confusion. What it looks like is written is,

[tex]13\ln (e^{x^2+2}) - e^{\ln (13x^2 + 26)} [/tex]

If this is correct, than what Pseudo and Halls have said stands.


To OP:
To reduce this, note that e and ln are inverse functions, as Pseudo and Halls have already mentioned. So, for any positive real number a, you get the following

[tex] \ln e^a = a [/tex]

[tex] e^{\ln a} = a[/tex]

Thus
[tex] \ln (e^{x^2 + 2}) = x^2 + 2 [/tex]
 
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  • #7
nocturnal said:
I think you're misreading the OP's post. Let's write this out in LaTex so there's no confusion. What it looks like is written is,

[tex]13\ln (e^{x^2+2}) - e^{\ln (13x^2 + 26)} [/tex]

If this is correct, than what Pseudo and Halls have said stands.To OP:
To reduce this, note that e and ln are inverse functions, as Pseudo and Halls have already mentioned. So, for any positive real number a, you get the following

[tex] \ln e^a = a [/tex]

[tex] e^{\ln a} = a[/tex]

Thus
[tex] \ln (e^{x^2 + 2}) = x^2 + 2 [/tex]

the function is 13ln (exp[(x^2+2)] - exp[ln(13x^2+26)])

NOT seperately as in (13ln[exp[(x^2+2)] ) - (exp[ln(13x^2+26)])

sorry if i was unclear, writing in laTex is too tedious for me
 
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FAQ: Simplifying Transcendental Functions

What are transcendental functions?

Transcendental functions are mathematical functions that cannot be expressed as a finite combination of algebraic functions. They are typically defined in terms of infinite series or integrals and include functions such as trigonometric, logarithmic, and exponential functions.

What is the significance of transcendental functions in scientific research?

Transcendental functions are essential in various fields of science, including physics, engineering, and mathematics. They are used to model and describe natural phenomena, such as the motion of waves, the growth of populations, and the behavior of electric circuits.

How are transcendental functions different from algebraic functions?

The main difference between transcendental and algebraic functions is that transcendental functions cannot be solved algebraically, while algebraic functions can be expressed using a finite number of algebraic operations (addition, subtraction, multiplication, division, and taking roots).

Can transcendental functions be graphed?

Yes, transcendental functions can be graphed using various mathematical software and calculators. However, since these functions are defined in terms of infinite series or integrals, their graphs are typically not smooth and may contain oscillations or asymptotes.

How are transcendental functions used in real-life applications?

Transcendental functions are used in many real-life applications, such as in engineering for designing structures and predicting their behavior, in physics for modeling the movement of objects and waves, in economics for analyzing market trends, and in biology for studying growth and decay processes.

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