SOLUTION: 2ˣ + 2^y = 40, x + y = 8 find x, y.

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Question 1208417: 2ˣ + 2^y = 40,
x + y = 8
find x, y.
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52750) About Me  (Show Source):
You can put this solution on YOUR website!
.
2^x + 2^y = 40,
x + y = 8
find x, y.
~~~~~~~~~~~~~~~~~~~~ 

Starting equations are 

    2^x + 2^y = 40,        (1)

    x + y = 8              (2)


Write equation (2) in equivalent form

    (x-4) + (y-4) = 0.     (3)


Introduce new variables 

    p = x-4,  q = y-4.     (4)


Then p + q = 0,  or  p = -q.    (5)

Equation (1) takes the form

    2^(p+4) + 2^(q+4) = 40,

or

    16*2^p + 16*2^q = 40,

     2*2^p + 2*2^q = 5.    (5)


Introduce new variable 

    u = 2^p.               (6)


Since p = -q, equation (5) takes the form

    2u + 2%2Fu = 5,    (7)

    2u^2 + 2 = 5u,

    2u^2 - 5u + 2 = 0,

    u%5B1%2C2%5D = %285+%2B-+sqrt%28%28-5%29%5E2+-4%2A2%2A2%29%29%2F%282%2A2%29 = %285+%2B-+sqrt%289%29%29%2F4 = %285+%2B-+3%29%2F4.


So, equation (7) has two roots  u%5B1%5D = %285+%2B+3%29%2F4 = 2,  u%5B2%5D = %285+-+3%29%2F4 = 1%2F2.


From (6), we have for p and q two possibilities

    p = log%282%2C+%282%29%29 = 1,  q = -p = -1,

or

    p = log%282%2C+%281%2F2%29%29 = - 1,  q = -p = 1.


It gives two possibilities for  x  and  y

    x = p+4 = 1+4 = 5,  y = q+4 = -1+4 = 3,   (8)

or


    x = p+4 = -1+4 = 3,  y = q+4 = 1+4 = 5.    (9)


ANSWER.  The given system has two solutions.

         One solution is  x= 5,  y= 3.

         The other solution is  x= 3,  y= 5.

Solved.

This problem is of the level of entrance examination on Math for renowned Ivy league universities.



Answer by MathTherapy(10549) About Me  (Show Source):
You can put this solution on YOUR website!
2ˣ + 2^y = 40, 
x + y = 8
find x, y. 

2x + 2y = 40 --- eq (i)
 x +  y = 8 ===> y = 8 - x  ---- eq (ii)
                matrix%281%2C3%2C+2%5Ex+%2B+2%5E%288+-+x%29%2C+%22=%22%2C+40%29 ----- Substituting 8 - x for y in eq (i)
                   matrix%281%2C3%2C+2%5Ex+%2B+%282%5E8%29%2F%282%5Ex%29%2C+%22=%22%2C+40%29 ---- Applying IDENTITY a(b - c) = a%5Eb%2Fa%5Ec to 2(8 - x)
          (2x)(2x) + 28 = 40(2x) --- Multiplying by LCD, 2x
 (2x)(2x) - 40(2x) + 28 = 0
     22x - 40(2x) + 256 = 0
   (2x)2 - 40(2x) + 256 = 0
         a2 - 40a + 256 = 0 ------ Substituting a for 2x
        (a - 32)(a - 8) = 0 -- Factoring the TRINOMIAL      
         a - 32 = 0     |   a - 8 = 0 ------ Setting each FACTOR equal to 0
              a = 32    |       a = 8
             2x = 32    |      2x = 8  ------ Back-substituting 2x for a
             2x = 25    |      2x = 23  ----- Substituting 25 for 32, and 23 for 8 
                 x = 5      OR        x = 3 ------- Bases are the same, and so are the EXPONENTS

   x = 5                                       x = 3
  y = 8 - 5 --- Substituting 5 for x in eq (ii)        y = 8 - 3 --- Substituting 3 for x in eq (ii)
  y = 3                                                 y = 5 

SOLUTION SETS: (x1, y1) = (5, 3)
               (x2, y2) = (3, 5)

You can do the check, if you so DESIRE!!