Question 1971
{{{e^(2ln(2x)) = 4}}}

Take logs of both sides, to remove the e power

2ln(2x) = ln4
2ln(2x) = ln{{{2^2}}}
2ln(2x) = 2ln2
ln(2x) = ln2

raise everything to power of e, to remove the logs

2x = 2
x=1

CHECK...always a good thing to do.

{{{e^(2ln(2))}}} plug this into your calculator and you get 4, so x=1 is correct.

or manually :-): {{{e^(ln(2^2))}}} --> {{{e^(ln(4))}}}

Now, {{{e^(ln(x))}}} just leaves x, so ours just leaves the 4 --> the answer

cheers
jon