how i show this? (1 Viewer)

[bos1234]

show that

=

 

[watatank]

it doesn't work for all x. try putting a few numbers in they don't equate

xe^x = x^{x + lnx}

consider LHS

LHS = xe^x

x = e^lnx

proof -----> (let y = x ---> lnx = lny ---> e^lny = x )

.'. LHS = e^lnx * e^x

= e^{x + lnx}

RHS = x^{x + lnx}

compare to LHS. its exactly the same bar the x and e being the base numbers.

so it seems it works when x = e? i tried that and that seemed to work quite fine...

afaik it only works for x = e.

 

[Slidey]

show that

=

The question is solve, I'm guessing:

e^x = x^(x-1+lnx)
e^x = e^([x-1+lnx].lnx)
x=(x-1+lnx)lnx
xlnx-lnx+(lnx)^2-x=0
x(lnx-1) + lnx(lnx-1)=0
(x+lnx)(lnx-1)=0
So the solutions are:
lnx=1 or when x=e
as well as any solutions to the transcendental equation:
lnx+x=0

Man that took too long. I miss maths... haha and I probably made mistakes.