Understanding Exponential and Natural Log Rules: Is this Simplification Correct?

Click For Summary

Homework Help Overview

The discussion revolves around the simplification of expressions involving exponential and logarithmic functions, specifically focusing on the equation e^{2(2x+ln2)} and its transformation into e^{4x}e^{2ln2}. Participants are exploring the validity of this simplification and its general applicability.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expansion of the expression and the application of exponential rules. Questions arise regarding the generality of the simplification and whether similar rules apply to other powers or functions.

Discussion Status

Some participants express initial confusion but then acknowledge understanding after further explanation. There is a recognition of the broader principles at play, although no explicit consensus is reached on the implications of these rules in different contexts.

Contextual Notes

Participants are navigating the nuances of exponential functions and logarithms, with some uncertainty about the definitions and contexts in which these rules apply. The discussion reflects a learning process influenced by a paper's mark scheme and personal interpretations.

thomas49th
Messages
645
Reaction score
0
Is it true that

[tex]e^{2(2x+ln2)} = e^{4x}e^{2ln2}[/tex]

I can't see how that is true? According to a paper mark scheme it is.

Can someone clarify
Thanks :)
 
Physics news on Phys.org
First open up the brackets:
[tex]e^{2(2x+\ln 2)} = e^{4x + 2 \ln 2}[/tex]
Then use a rule for [tex]e^{a + b}[/tex] which you (should) know.

Actually, you can even simplify it further to [itex]4 e^{4x}[/itex].
 
is this the same with all powers or just the exponetial function... hang on, is it a function?
 
Well, the first simplification (the one you asked about) works in general. If x is any number and a and b are two expressions, then
[tex]x^{a + b} = x^a x^b[/tex]
so in particular it works for x = e, a = 4x and b = 2 ln(2).

The further simplification I spoke about just works because ln[..] is the inverse of exp[..] = e^[...]
 
yeah i can see the second one.
The first one makes sense now. Yes. I just didn't see it with the 'e'

cheerz :)
 
So the lesson to be learnt, perhaps, is that
"[tex]e^x[/tex]"​
can be viewed both as the function "exp" evaluated in (some number) x, or as the number [itex]e \approx 2,7\cdots[/itex] raised to the power (some number) x and that how you view it depends on the context (e.g. when differentiating it, one should view it as a function; when using simplification rules like here it's easier to just view it as an exponentiation).
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 44 ·
2
Replies
44
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K