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Exponetial and natural log

  • Thread starter thomas49th
  • Start date
  • #1
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Is it true that

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

I cant see how that is true? According to a paper mark scheme it is.

Can someone clarify
Thanks :)
 

Answers and Replies

  • #2
CompuChip
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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].
 
  • #3
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is this the same with all powers or just the exponetial function... hang on, is it a function?
 
  • #4
CompuChip
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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^[...]
 
  • #5
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yeah i can see the second one.
The first one makes sense now. Yes. I just didn't see it with the 'e'

cheerz :)
 
  • #6
CompuChip
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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).
 

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