MM2002
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I don’t see how they got from line 1 to line 2! What’s the rule called ?! Cheers
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MM2002
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RDKGames
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(Original post by MM2002)
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Inverse functions.

e^x and ln x are inverse functions of each other.

So f(f^{-1}(x)) = e^{\ln x} = x
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davros
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(Original post by MM2002)
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Do you mean the very first equallity?
We always have z = e^{ln z} - that just expresses the fact that e^w and ln w are inverse functions
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MM2002
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(Original post by RDKGames)
Inverse functions.

e^x and ln x are inverse functions of each other.

So f(f^{-1}(x)) = e^{\ln x} = x
I still don’t get it ! Could u pls recommend a video
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MM2002
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(Original post by MM2002)
I still don’t get it ! Could u pls recommend a video
First step and last step are confusing me
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davros
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(Original post by MM2002)
First step and last step are confusing me
Why do you need a "video"?

Raising e to the power of something and taking the (natural) logarithm are inverse functions (you have to be a bit careful post A level because log can have multiple values, but we don't need to worry about that here).

So if w = ln y then y = e^w

This is basically a definition (at A level). So if we know another form for y we can just substitute it into this relationship

e.g. if y = 3^x then  ln y = ln (3^x) and y = e^{ln y} = e^{ln(3^x)}
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old_engineer
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(Original post by MM2002)
First step and last step are confusing me
We can break the first step down into two stages.
1) Take the natural log of both sides.
\ln y=\ln(a^{kx})=kx\ln a=(k\ln a)x

2) Take the exponent of both sides
y=e^{(k\ln a)x}
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MM2002
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(Original post by old_engineer)
We can break the first step down into two stages.
1) Take the natural log of both sides.
\ln y=\ln(a^{kx})=kx\ln a=(k\ln a)x

2) Take the exponent of both sides
y=e^{(k\ln a)x}
Thank you! I get that part now. Last line I don't see how the e is lost
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RDKGames
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(Original post by MM2002)
Thank you! I get that part now. Last line I don't see how the e is lost
It's not so much 'lost', it reduces down to a^{kx} with this exact relation.
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