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Solve the equation $$8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}$$ The given equation is equivalent to $$2^{3x}+\dfrac{12}{2^x}=1+\dfrac{8}{2^{3x}}+6\cdot2^x$$ If we put $a:=2^x>0$, the equation becomes $$a^3+\dfrac{12}{a}=1+\dfrac{8}{a^3}+6a$$ which is $$a^6-6a^4-a^3+12a^2-8=0$$ The LHS factors as $(a+1)(a-2)(a^4+a^3-3a^2-2a+4)$, which is in no case obvious. Let's say that we find the roots $1$ and $-2$, then how do we show that $(a^4+a^3-3a^2-2a+4)$ does not factor any more? Taking these into consideration, I believe there is an another approach. Any ideas would be appreciated.

kormoran
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  • "how do we show that (degree-4 polynomial) does not factor any more?" Every complex polynomial can factor into linear components with complex coefficients as per the fundamental theorem of algebra, however not necessarily into factors with integer coefficients. There does exist a generic formula for quartics and below using "elementary" functions such as n'th roots and the like, however no such generic formula exists for quintics or higher (though special cases may exist). Actually performing such a factoring however can be very difficult. – JMoravitz Jan 16 '23 at 16:45
  • the remaining factor $a^4 + a^3 - 3 a^2 -2a + 4$ has no real roots: it is always positive. – Will Jagy Jan 16 '23 at 16:45
  • Hint : let $$a=2^x$$ – Moustapha_M_I Jan 16 '23 at 17:34

2 Answers2

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Hint

Let's start from

$$a^3+\dfrac{12}{a}=1+\dfrac{8}{a^3}+6a$$

Now rewrite it as

$$\left[a^3-\left(\frac2a\right)^3\right]-6\left[a-\frac 2a\right]-1=0$$ $$\left(a-\frac2a\right)\left[a^2+\frac{4}{a^2}-4\right]-1=0$$ $$\left(a-\frac2a\right)^3-1=0$$

Can you finish?

Arnaldo
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Letting $y=2^{x-\frac12}>0$

$$\begin{align}8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}&\iff y^3+\frac3y=\frac1{2\sqrt2}+\frac1{y^3}+3y \\&\iff\left(y-\frac1y\right)^3=\left(\frac1{\sqrt2}\right)^3\\&\iff\dots\\&\iff y=\sqrt2\\&\iff x=1.\end{align}$$

Anne Bauval
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