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Solve \((2 \times 3^x)^3 +(9^x - 3)^3 = (9^x + 2 \times 3^x - 3)^3\)

 Jul 7, 2020
 #1
avatar+783 
+2

Never mind.

 Jul 7, 2020
edited by gwenspooner85  Jul 8, 2020
 #2
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-1

No cussing here, but you (swearing edited out)  big time!

 

X=0 or X=1/2

Guest Jul 7, 2020
edited by Melody  Jul 8, 2020
 #3
avatar+783 
+1

What the heck is wrong with you? Why must you use such explicit words instead of saying something normally?! So what if I got it wrong?

gwenspooner85  Jul 7, 2020
edited by gwenspooner85  Jul 7, 2020
 #4
avatar+28021 
+1

Perhaps our linguistically challenged 'Guest' can explain their answer?

 

WolframAlpha finds   x = 1/ 2    or  0          Can Guest provide the solution......or just the answer (which really helps no one) ????

 Jul 7, 2020
 #7
avatar+4599 
0

A keen obseravation is to write \(y=3^x\) and \(y^2=9^x\).

After some simplifying and expainding, we get 2 negative and 2 positive soluions and y=0.

 

The two positive solutions are y=\(\sqrt{3}\) and y=\(1\), thus the corresponding values of x are 0 and 1/2.

 Jul 8, 2020
edited by tertre  Jul 8, 2020
 #8
avatar+111566 
+1

\((2 \times 3^x)^3 +(9^x - 3)^3 = (9^x + 2 \times 3^x - 3)^3\\ let\;\; y=3^x\\ (2y)^3 +(y^2 - 3)^3 = (y^2 + 2y - 3)^3\\ LHS=y^6-9y^4+8y^3+27y-27\\ RHS=y^6+6y^5+3y^4-28y^3-9y^2+54y-27\\ so\\ -9y^4+8y^3+27y=6y^5+3y^4-28y^3-9y^2+54y\\ \color{red}y=0\;\;\text{could lead to one solution}\qquad \color{black}or\\ -9y^3+8y^2+27=6y^4+3y^3-28y^2-9y+54\\ 6y^4+12y^3-36y^2-36y+54=0\\ y^4+2y^3-6y^2-6y+9=0 \)

 

\(consider\\ f(y)=y^4+2y^3-6y^2-6y+9\\ \text{Any integer roots will be factors of 9 so they could be}\pm1,\;\;\pm3,\;\;\pm9\\ f(1)=0\;\;\text{therefore it is divisible by }(y-1)\\ f(-3)=0\;\;\text{therefore it is divisible by }(y+3)\\ \text{by division I found another factor to be}\;\; (y^2-3)\\ so\\ f(y)=(y-1)(y+3)(y^2-3)\\ (y-1)(y+3)(y-\sqrt3)(y+\sqrt3)=0\\ \text{So I have}\\ y=0,1,-3,\sqrt3, -\sqrt3\\~\\ y=3^x\qquad x=log_3y\qquad \text{y has to be greater than 0 so}\\ y=1,\sqrt3\\ x=0\;\;or\;\;0.5 \)

 

 

 

Latex:

(2 \times 3^x)^3 +(9^x - 3)^3 = (9^x + 2 \times 3^x - 3)^3\\
let\;\; y=3^x\\
(2y)^3 +(y^2 - 3)^3 = (y^2 + 2y - 3)^3\\
LHS=y^6-9y^4+8y^3+27y-27\\
RHS=y^6+6y^5+3y^4-28y^3-9y^2+54y-27\\
so\\  -9y^4+8y^3+27y=6y^5+3y^4-28y^3-9y^2+54y\\
\color{red}y=0\;\;\text{could lead to one solution}\qquad \color{black}or\\
-9y^3+8y^2+27=6y^4+3y^3-28y^2-9y+54\\
6y^4+12y^3-36y^2-36y+54=0\\
y^4+2y^3-6y^2-6y+9=0

 

consider\\
f(y)=y^4+2y^3-6y^2-6y+9\\
\text{Any integer roots will be factors of 9 so they could be}\pm1,\;\;\pm3,\;\;\pm9\\
f(1)=0\;\;\text{therefore it is divisible by }(y-1)\\
f(-3)=0\;\;\text{therefore it is divisible by }(y+3)\\
\text{by division I found another factor to be}\;\; (y^2-3)\\
so\\
f(y)=(y-1)(y+3)(y^2-3)\\
(y-1)(y+3)(y-\sqrt3)(y+\sqrt3)=0\\
\text{So I have}\\
y=0,1,-3,\sqrt3, -\sqrt3\\~\\
y=3^x\qquad x=log_3y\qquad \text{y has to be greater than 0  so}\\
y=1,\sqrt3\\
x=0\;\;or\;\;0.5

 Jul 8, 2020
 #10
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0

With one keen observation and a detailed answer for the reals, there is no need for another …unless someone wants to solve for one or more of the complex answers.

 

Would this post have received this amount of attention without the troll post comments?  Nyet. This post likely would sit here in perpetuity with a wrong answer. 

 Jul 8, 2020

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