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Find all real x that satisfy (2^x - 4)^3 + (4^x - 2)^3 = (4^x + 2^x - 6)^3

 Jan 25, 2018
 #1
avatar
+2

Solve for x over the real numbers:
(2^x - 4)^3 + (4^x - 2)^3 = (-6 + 2^x + 4^x)^3

After a few expansions and re-arrangements, we have:

The left hand side factors into a product with five terms:
-3 (2^x - 4) (2^x - 2) (2^x + 3) (2^(2 x) - 2) = 0

Divide both sides by -3:
(2^x - 4) (2^x - 2) (2^x + 3) (2^(2 x) - 2) = 0

Split into four equations:
2^x - 4 = 0 or 2^x - 2 = 0 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Add 4 to both sides:
2^x = 4 or 2^x - 2 = 0 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

4 = 2^2:
2^x = 2^2 or 2^x - 2 = 0 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Equate exponents of 2 on both sides:
x = 2 or 2^x - 2 = 0 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Add 2 to both sides:
x = 2 or 2^x = 2 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

2 = 2^1:
x = 2 or 2^x = 2^1 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Equate exponents of 2 on both sides:
x = 2 or x = 1 or 2^x + 3 = 0 or 2^(2 x) - 2 = 0

Subtract 3 from both sides:
x = 2 or x = 1 or 2^x = -3 or 2^(2 x) - 2 = 0

2^x = -3 has no solution since for all z element R, 2^z>0 and -3<0:
x = 2 or x = 1 or 2^(2 x) - 2 = 0

Add 2 to both sides:
x = 2 or x = 1 or 2^(2 x) = 2

Take the logarithm base 2 of both sides:
x = 2 or x = 1 or 2 x = 1

Divide both sides by 2:
x = 2            or            x = 1         or              x = 1/2

 Jan 25, 2018
 #2
avatar+130458 
+2

(2^x - 4)^3 + (4^x - 2)^3 = (4^x + 2^x - 6)^3

 

Let  a  = 2^x    let  a^2  =  (2^x)^2  = (2^2)^x  =  4^x

 

So we have

 

( a - 4)^3  + ( a^2 - 2)^3  =  ( a^2 + a - 6)^3

 

(a - 4)^3  + (a^2 - 2)^3  =  (a^2 + a - 6)^3      write the first two terms as a sum of cubes

 

[ (a - 4) +( a^2 - 2)]  [  (a - 4)^2 - (a - 4)(a^2 - 2) + (a^2 - 2)^2]  =  (a^2 + a - 6)^3

 

[ a^2 + a - 6 ]   [ (a - 4)^2 - ( a- 4) (a^2 - 2) + (a^2 - 2)^2 ]  =  [ a^2 + a - 6 ]^3

 

 a^6 - 6 a^4 + a^3 + 48 a - 72   = a^6 + 3 a^5 - 15 a^4 - 35 a^3 + 90 a^2 + 108 a - 216

 

3a^5  - 9a^4 - 36a^3 + 90a^2 + 60a - 144  = 0  this factors as

 

3 (a - 4) (a - 2) (a + 3) (a^2 - 2) = 0

 

Since a = 2^x....then only the positive roots need be considered

 

So

 

a = 4        a   = 2        and    a  = √2

 

So

 

2^x  =  4    ⇒  x  = 2

2^x  =  2  ⇒  x = 1

2^x  = √2   ⇒  x = 1/2 

 

 

cool cool cool

 Jan 25, 2018
 #3
avatar+26396 
+2

Find all real x that satisfy (2^x - 4)^3 + (4^x - 2)^3 = (4^x + 2^x - 6)^3

 

(2x4)3+(4x2)3=(4x+2x6)3We substitute: a=2x, b=4x, c=4, d=2(a+c)3+(b+d)3=(a+b+c+d)3a3+3ac(a+c)+c3+b3+3bd(b+d)+d3=a3+b3+c3+d3+6abc+6abd+6acd+6bcd+3a2b+3a2c+3a2d+3ab2+3ac2+3ad2+3b2c+3b2d+3bc2+3bd2+3c2d+3cd2a3+b3+c3+d3+3a2c+3ac2+3b2d+3bd2=a3+b3+c3+d3+6abc+6abd+6acd+6bcd+3a2b+3a2c+3a2d+3ab2+3ac2+3ad2+3b2c+3b2d+3bc2+3bd2+3c2d+3cd20=6abc+6abd+6acd+6bcd+3a2b+3a2d+3ab2+3ad2+3b2c+3bc2+3c2d+3cd2 | :30=2abc+2abd+2acd+2bcd+a2b+a2d+ab2+ad2+b2c+bc2+c2d+cd20=a2(b+d)+b2(a+c)+c2(b+d)+d2(a+c)+2bd(a+c)+2ac(b+d)0=(b+d)(a2+2ac+c2)+(a+c)(b2+bd+d2)0=(b+d)(a+c)2+(a+c)(b+d)20=(b+d)(a+c)(a+b+c+d)

 

Solution:

b+d=04x2=04x=222x=212x=1x1=12

 

a+c=02x4=02x=42x=22x=2x2=2

 

a+b+c+d=02x+4x6=0(2x2)(2x+3)=02x2=02x=21x=1x3=12x+3=02x=3xlog(2)=log(3)| no solution, log(3) complex 

 

laugh

 Jan 25, 2018

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