<p>So, a math problem, not from SATs or anything, one my friend posed me. </p>
<p>2^x = 2x.</p>
<p>Solve. Yes, x = 1 or 2. Solve algebraically. (sp?)</p>
<p>This is going to be unsatisfying, but you can’t find exact solutions algebraically - you have to approximate through one of various methods. A good idea would be to graph and check. I think (I haven’t actually graphed =p ) that there are only two solutions, and since you have them, that’s it.</p>
<p>The two answers are 23.56, and -9.66671 respectively.</p>
<p>Ah. Actually, tetrahedr0n, that’s not too bad. It’s better than me somehow completely misusing basic logarithms and there being a simple solution, anyway 
Can you solve it with limits/calculus? Newton’s method? I’m not sure how that would work, but my friend’s school’s AP calculus teacher mentioned limits. </p>
<p>GoldShadow: … no, they aren’t 
Check! It’s 2 to the xth power, I’m not sure what you did.</p>
<p>What I did was post two random numbers off the top of my head, heh.</p>
<p>2^x=2x
log2^x=log2x
xlog2=log2x…</p>
<p>DAMN!@!!! There’s gotta be a way to do this **** with logarithms. I will declare anyone who can solve this problem with logarithms a math god. I believe I started on the right path…maybe…:mad:</p>
<p>u really can’t i don’t think. Newton’s method is approximation and u really can’t do it without a calculator b/c it takes really like 5-6 steps for really accurate stuff. But even still there will not be an exact number. i’m thinking mclaurin series and approximating somehow.</p>
<p>according to logs/lns you can’t really do it…for example…make the equation 2^x - 2x = 0</p>
<p>now…set it to logs or lns whichever one you like best</p>
<p>2lnx - ln2 - lnx = ln0 which…ln0=no solution…hm</p>
<p>nahrafsfa, isn’t it xln2, not 2lnx?</p>
<p>whoops my bad</p>
<p>yeah same thing…it’ll still end up being unsolvable…i’ve been using all the log rules i know and at one point i ended up thinking i got it and it gave me the EXACT same thing i started out with…it’s a samsara of logarithms…no solution dammit</p>