- #1

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[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 :)

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- Thread starter thomas49th
- Start date

- #1

- 655

- 0

[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

CompuChip

Science Advisor

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[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

- 655

- 0

is this the same with all powers or just the exponetial function... hang on, is it a function?

- #4

CompuChip

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[tex]x^{a + b} = x^a x^b[/tex]

so in particular it works for

The further simplification I spoke about just works because ln[..] is the inverse of exp[..] = e^[...]

- #5

- 655

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The first one makes sense now. Yes. I just didn't see it with the 'e'

cheerz :)

- #6

CompuChip

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"[tex]e^x[/tex]"

can be viewed both as the
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