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Natural Logging...
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\[e ^{\ln x}\]
If I natural log that... the x would come down right?
yes
Or would lnx all come down?
I'm thinking lnx...
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y = e^(lnx) lny = ln(e^(lnx)) lny = lnx ln(e) lny = lnx * 1 lny = lnx
\(e^x\) and \(\ln x\) are inverse functions, so \(e^{\ln x} = x\)
\[\large f(f^{-1}(x))=f^{-1}(f(x))=x\]
ok
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