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Exponetial and natural log

  1. May 30, 2008 #1
    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 :)
  2. jcsd
  3. May 30, 2008 #2


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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].
  4. May 30, 2008 #3
    is this the same with all powers or just the exponetial function... hang on, is it a function?
  5. May 30, 2008 #4


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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^[...]
  6. May 30, 2008 #5
    yeah i can see the second one.
    The first one makes sense now. Yes. I just didn't see it with the 'e'

    cheerz :)
  7. Jun 1, 2008 #6


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    So the lesson to be learnt, perhaps, is that
    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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