Wierd. The answer is obviously x =  2 but I can't seem to solve it using logarithms or anything like that.

Hmm. I don't really know what I'm doing here, but I'll give it a go.

2^x = 2x
2^(x-1) = x
x-1 = lg(2) x

...and I don't know after that. Bleh.

x = 1 would also be an answer.

Yeah I tried logs and changes of bases and etc but it just simplifies to the original equation.

The only answers are x = 1 and x = 2, which can be proved once you do "find" them...

Hi.
I tried a thing like this  (but I think that it's the same thing)
Let ln log base e
2^x=2x
ln(2^x)=ln2+lnx
xln2=ln2+lnx
x=1+ln(x)/ln(2)
and so

x=1+Log    (x)
               2

at the end of post#5 I had  x-1 = x/2, but is this allowed since
I took the inverse of the functions?

We have two answers:
x=1 and x=2.

And we'll proof that there doesn't exist line that intersects with 2^i more than 2 times. That's because
ln x means natural logaritm of x
(2^i)'=2^i(ln2)=2^iC
That means that (2^i)' is a monotonic growing function so (2^i) is "convex".

Could somebody proof the upper using analisis?
I'll try solving it but i'm not so good at analisis.