+0  
 
+1
46
1
avatar+70 

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

 Jan 27, 2021
 #1
avatar
0

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

(2^x - 4)^3 + (4^x - 2)^3 = -72 + 2^(3 x) + 3 2^(x + 4) - 3 2^(2 x + 2) - 3 2^(4 x + 1) + 4^(3 x) + 3 4^(x + 1):
-72 + 2^(3 x) + 3 2^(x + 4) - 3 2^(2 x + 2) - 3 2^(4 x + 1) + 4^(3 x) + 3 4^(x + 1) = (-6 + 2^x + 4^x)^3

(-6 + 2^x + 4^x)^3 = -216 + 2^(3 x) + 2^(4 x) + 2^(5 x) + 9 2^(x + 2) + 9 2^(x + 3) - 3 2^(2 x + 1) - 3 2^(2 x + 2) + 9 2^(2 x + 3) - 3 2^(3 x + 2) - 3 2^(3 x + 3) - 2^(4 x + 2) + 2^(5 x + 1) + 4^(3 x) + 9 4^(x + 1) - 3 4^(2 x + 1):
-72 + 2^(3 x) + 3 2^(x + 4) - 3 2^(2 x + 2) - 3 2^(4 x + 1) + 4^(3 x) + 3 4^(x + 1) = -216 + 2^(3 x) + 2^(4 x) + 2^(5 x) + 9 2^(x + 2) + 9 2^(x + 3) - 3 2^(2 x + 1) - 3 2^(2 x + 2) + 9 2^(2 x + 3) - 3 2^(3 x + 2) - 3 2^(3 x + 3) - 2^(4 x + 2) + 2^(5 x + 1) + 4^(3 x) + 9 4^(x + 1) - 3 4^(2 x + 1)

Subtract -216 + 2^(3 x) + 2^(4 x) + 2^(5 x) + 9 2^(x + 2) + 9 2^(x + 3) - 3 2^(2 x + 1) - 3 2^(2 x + 2) + 9 2^(2 x + 3) - 3 2^(3 x + 2) - 3 2^(3 x + 3) - 2^(4 x + 2) + 2^(5 x + 1) + 4^(3 x) + 9 4^(x + 1) - 3 4^(2 x + 1) from both sides:
144 - 2^(4 x) - 2^(5 x) - 9 2^(x + 2) - 9 2^(x + 3) + 3 2^(x + 4) + 3 2^(2 x + 1) - 3 2^(2 x + 5) + 3 2^(3 x + 2) + 3 2^(3 x + 3) - 3 2^(4 x + 1) + 2^(4 x + 2) - 2^(5 x + 1) + 3 4^(2 x + 1) = 0

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 27, 2021

69 Online Users

avatar
avatar
avatar
avatar
avatar