Find all real x that satisfy (2^x - 4)^3 + (4^x - 2)^3 = (4^x + 2^x - 6)^3
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
(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
Find all real x that satisfy (2^x - 4)^3 + (4^x - 2)^3 = (4^x + 2^x - 6)^3
(2x−4)3+(4x−2)3=(4x+2x−6)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+3cd2⧸a3+⧸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=04x−2=04x=222x=212x=1x1=12
a+c=02x−4=02x=42x=22x=2x2=2
a+b+c+d=02x+4x−6=0(2x−2)(2x+3)=02x−2=02x=21x=1x3=12x+3=02x=−3xlog(2)=log(−3)| no solution, log(−3) complex