Calculate this and get a FullInfoAccount

[BACKABACKA]

id have to say, 5

[maclone]

42.
I don't even understand what to do there.

[ReidE96]

Thanks to all my working, which I'm now sure contains an error (see if you can find it, anyone?), I'm going to guess the wrong answer of 1.5

[camicio]

Here's how far I got:
A is the point between 0 and 1 where the function crosses the x axis and B is the point between 0 and 1 where the function crosses the x=1 axis.
A has coordinates (exp(-a),0) (since 0=ln(x)+a)
B has coordniate (1,....)
So you have to integrate ln(x)+a between exp(-a) and 1, which has to be equal to 1/2, giving you 3/2=exp(-a)+a, which I don't know how to solve (you can only do it numerally)

[camicio]

It's between 1.1982904373156675 and 1.1982904373156676, imo. I'm not sure, that's using math.h's exp() function.

[Y R U A NUB ?]

Thanks to all my working, which I'm now sure contains an error (see if you can find it, anyone?), I'm going to guess the wrong answer of 1.5

You are indeed a scary/genious creature :eek4dance:

[blubb12345]

Find a value of a such that the area beneath the curve is 0.5
Area beneath f(x) is integral of f(x)


wrong :P (integral of ln(X) is x*ln(x)-x (x>0))

|||||||||
vvvvvvvv
Integral of f(x) = integral of ln(x) + ax +c
=xlnx + ax + c - integral of 1
=xlnx +(a-1)x + c

Upper limit is 1 always, lower limit is... hmmm....

1ln(1) + a - 1 - (xlnx + (a-1)x) = 0.5
a - xlnx - ax + x = 1.5
a(1-x) = 1.5 + x(lnx - 1) = (1.5 + x(lnx-1))/(1-x)

Find the lower limit of x, plug it in, get your answer!

Edit: Thought a little more.

Lower limit is where f(x) = 0 so
ln(x) = -a
x = 1/(e^a)

1ln(1) + a - 1 - (-a/(e^a) + (a-1)/(e^a)) = 0.5
a + a/(e^a) - (a-1)/(e^a) = 1.5
a + e^-a = 1.5
a + 1/(e^a) = 1.5
ae^a + 1 = 1.5e^a
(a - 1.5)e^a = -1

Yeah, I cba now, dinner's ready Enjoy folks!

Edit again: Back from dinner, and seeing as noone has it yet I'll push on.

a + e^-a = 1.5
e^-a = 1.5 - a

We know ln(x) + a = 0 at x = e^(-a)
But we also know e^(-a) = 1.5 - a
So ln(1.5 - a) + a = 0
ln(1.5)/ln(a) = -a
ln(1.5) = -aln(a)
ln(1.5) = ln(a^-a)
a^-a = 1.5
a^a = 2/3

Ok, that's just great.

Lemme try again, from (a - 1.5)e^a = -1
ln of both sides: ln((a-1.5)e^a) = -1
ln(a-1.5) + ln(e^a) = -1
ln(a-1.5) = -a - 1
a = 1.5 + e^(-a - 1)
a = 1.5 + (e^-a)/(e^-1)
a = 1.5 + (e^1)/(e^a)
a = 1.5 + e*e^(-a)

But e^-a = 1.5 - a, so

a = 1.5 + e(1.5 - a)
a = 1.5(1+e) - ea
a(1+e) = 1.5(1+e)
a = 1.5

WOO! Now, common sense tells me something's wrong there. But I don't care

i am sorry, its wrong
and the rest is also wrong

[Illidan_000]

wow ur theory is so hard XD

[Wazabara]

for x= 1.648721271
and a=0!!


God...this is stupid^^

[skatrdie94]

9001, i win.

[Mitron]

omfg only einstein can solve this shit o.o

[_DecpticoN_]

f(½) = 0

Solve

ln(½) + a = 0
−ln(2) + a = 0
a = ln(2)

[Judas911]

A passes through .707 and 0
it should be a = .3465735903

[alj03]

It = 1337.

[Mercer]

f(1/2) = 0 Solve ln(1/2) + a = 0-ln(2) + a = 0a = ln(2) therefore a = 0.693147180559945