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Jun 12th, 2006, 11:36 AM
#1
Thread Starter
New Member
Proof: "lnx ≠ x" (please check!)
Hi,
i'm doing AS Further Maths so if i've made assumptions that are over ambiguos or done anything wrong that i haven't realised yet then please correct me.
Prove that lnx ≠ x:
I thought the best way was proof by contradiction (alternative methods would be hugely appreciated) so;
If this is true then y = lnx would intersect the line y = x.
As y = e^x is a reflection of y = lnx and visa versa,
Then y = e^x would intersect y = lnx.
So when does:
lnx = e^x (apply e to both sides)
x = e^e^x (differentiate)
1 = e^x X e^e^x (ln both sides)
0 = ln(e^x X e^e^x)
0 = x + e^x
e^x = -x (Which is a contradiction)
So if lnx did = e^x then we would have a contradiction thus lnx ≠ e^x and thus there are no values of x such that lnx = x. End of Proof...I hope
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