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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
+1

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+111331 
+1

(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+24995 
+1

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

 

\(\begin{array}{|lrcll|} \hline (2^x - 4)^3 + (4^x - 2)^3 = (4^x + 2^x - 6)^3 \\ \qquad\text{We substitute:} ~ a = 2^x,~ b = 4^x,~ c = -4,~ d = -2 \\ (a +c)^3 + (b +d)^3 = (a+b+c+d)^3 \\ \hline \end{array}\\ \small{ \begin{array}{|lrcll|} \hline a^3+3ac(a+c)+c^3+b^3+3bd(b+d)+d^3 & =& a^3 + b^3 + c^3+ d^3 \\ && + 6 a b c + 6 a b d + 6 a c d + 6 b c d \\ & & + 3 a^2 b + 3 a^2 c + 3 a^2 d + 3 a b^2 \\ & & + 3 a c^2 + 3 a d^2 + 3 b^2 c + 3 b^2 d \\ & & + 3 b c^2 + 3 b d^2 + 3 c^2 d + 3 c d^2 \\ \not{a^3}+\not{b^3}+\not{c^3}+\not{d^3}+\not{3a^2c}+\not{3ac^2}+\not{3b^2d}+\not{3bd^2} & =& \not{a^3} + \not{b^3} + \not{c^3}+ \not{d^3} \\ && + 6 a b c + 6 a b d + 6 a c d + 6 b c d \\ & & + 3 a^2 b + \not{3 a^2 c} + 3 a^2 d + 3 a b^2 \\ & & + \not{3 a c^2} + 3 a d^2 + 3 b^2 c + \not{3 b^2 d} \\ & & + 3 b c^2 + \not{3 b d^2} + 3 c^2 d + 3 c d^2 \\ \hline \end{array}}\\ \small{ \begin{array}{|rcll|} \hline 0 &=& 6 a b c + 6 a b d + 6 a c d + 6 b c d + 3 a^2 b + 3 a^2 d + 3 a b^2 + 3 a d^2 + 3 b^2 c + 3 b c^2 + 3 c^2 d + 3 c d^2 ~ | ~: 3 \\ 0 &=& 2 a b c + 2 a b d + 2 a c d + 2 b c d + a^2 b + a^2 d + a b^2 + a d^2 + b^2 c + b c^2 + c^2 d + c d^2 \\ 0 &=& a^2(b+d)+b^2(a+c)+c^2(b+d)+d^2(a+c)+2bd(a+c)+2ac(b+d) \\ 0 &=& (b+d)(a^2+2ac+c^2)+(a+c)(b^2+bd+d^2) \\ 0 &=& (b+d)(a+c)^2+(a+c)(b+d)^2 \\ 0 &=& (b+d)(a+c)(a+b+c+d) \\ \hline \end{array}}\)

 

Solution:

\(\begin{array}{|rcll|} \hline \mathbf{b+d} &\mathbf{=}& \mathbf{0} \\ 4^x-2 &=& 0 \\ 4^x &=& 2 \\ 2^{2x} &=& 2^1 \\ 2x &=& 1 \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{a+c} &\mathbf{=}& \mathbf{0} \\ 2^x-4 &=& 0 \\ 2^x &=& 4 \\ 2^{x} &=& 2^2 \\ x &=& 2 \\ \mathbf{x_2} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{a+b+c+d} &\mathbf{=}& \mathbf{0} \\ 2^x + 4^x-6 &=& 0 \\ (2^x-2)(2^x+3) &=& 0 \\ \hline 2^x-2 &=& 0 \\ 2^x &=& 2^1 \\ x &=& 1 \\ \mathbf{x_3} &\mathbf{=}& \mathbf{1} \\\\ 2^x+3 &=& 0 \\ 2^x &=& -3 \\ x\log(2) &=& \log(-3) \quad & |\quad \text{ no solution, $log(-3)$ complex } \\ \hline \end{array}\)

 

laugh

 Jan 25, 2018

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