wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> easy >> Solve For X.
        
(Message started by: rloginunix on Oct 25^th, 2014, 11:09am)

Title: Solve For X.
Post by rloginunix on Oct 25^th, 2014, 11:09am

Solve For X:

(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(2 +http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3))^x + (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(2 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3))^x = 2^x

Title: Re: Solve For X.
Post by dudiobugtron on Oct 25^th, 2014, 7:27pm

lol, I think the answer is [hide]x = 2[/hide]

This is a cool puzzle.

Title: Re: Solve For X.
Post by rloginunix on Oct 26^th, 2014, 8:08am

The answer is correct but what about the proof that there are no other solutions? That's the interesting part.

Title: Re: Solve For X.
Post by dudiobugtron on Oct 26^th, 2014, 2:02pm

Graphically, we can see that [hide](http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(2 +http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3))^x + (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(2 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3))^x - 2^x is always decreasing. You can also differentiate it and graph that to show that the gradient is always negative.[/hide]

I imagine that's not the interesting way of solving it you were talking about though!

Title: Re: Solve For X.
Post by rloginunix on Oct 27^th, 2014, 7:27am

The approach I have in mind is rather simple - it has a tri[hide]g[/hide] to it.

Start by [hide]dividing both sides of the equation by 2^x[/hide].

Title: Re: Solve For X.
Post by towr on Oct 27^th, 2014, 9:45am

[hide]And then notice one side is strictly monotonically increasing and the other side constant and therefore only cross in one point[/hide]?

Title: Re: Solve For X.
Post by rloginunix on Oct 27^th, 2014, 11:37am

That is one way of doing it. The other, a bit more explicit, is to recall that you already know what x is. Keeping the [hide]trigonometric[/hide] clue in mind we have: [hide]something squared plus something squared equals one[/hide] ...

Title: Re: Solve For X.
Post by SWF on Oct 27^th, 2014, 7:59pm

I see what rloginunix is looking for: [hide]cos(pi/12)^x+sin(pii/12)^x=1, and left hand side is a decreasing function of x, so 1 solution[/hide].

Here is another way: let y=x/2:
(2+sqrt(3))^y + (2-sqrt(3))^y = 4^y
Divide by 2+sqrt(3) and rearrange to get:
1+z^y = (1 + z)^y
where z is 7-4*sqrt(3), and is between 0 and 1.
Left hand side decreases with y and right hand side increases with y, so 1 solution (y=1).

Title: Re: Solve For X.
Post by rloginunix on Oct 28^th, 2014, 8:36am

You got it, SWF.

[hide]The argument to Sin and Cos is either http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/12, as you have it, or, if you switch the substitutions around, 5http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/12[/hide] but the numbers come out the same.

Title: Re: Solve For X.
Post by dudiobugtron on Oct 28^th, 2014, 5:19pm

There's a more general case to this, although it uses towr's/SWF's idea rather than a trig identity!

-----------------------

For any numbers a, b and c with 0 < a,b < c

a^x + b^x = c^x

has at most one solution for x.

proof: divide both sides by c^x

Then you have (a/c)^x + (b/c)^x = 1

a/c and b/c are both less than 1, so the left hand side is strictly monotonically decreasing. Therefore, there's at most one solution.

(It's pretty easy to show using the intermediate value theorem that there will be at least one solution; that's an exercise left to the reader though!)