Simplifying Transcendental Functions

[MercuryRising]

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

 

[Pseudo Statistic]

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'. :)

 

[HallsofIvy]

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'.

 

[daveb]

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.

 

[MercuryRising]

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..

 

[nocturnal]

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,

$$13\ln (e^{x^2+2}) - e^{\ln (13x^2 + 26)}$$

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

$$\ln e^a = a$$

$$e^{\ln a} = a$$

Thus
$$\ln (e^{x^2 + 2}) = x^2 + 2$$

 

[MercuryRising]

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,

$$13\ln (e^{x^2+2}) - e^{\ln (13x^2 + 26)}$$

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

$$\ln e^a = a$$

$$e^{\ln a} = a$$

Thus
$$\ln (e^{x^2 + 2}) = x^2 + 2$$

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

NOT separately 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