Exponential equations (1 Viewer)

jathu123 · Mar 6, 2017

A straight forward method to find the real solutions:

$(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3 \\ $ Let $a\equiv 2^x-4, $ and $b\equiv 4^x-2 \\ \therefore a^3+b^3=(a+b)^3 \\ (a+b)(a^2-ab+b^2)=(a+b)^3 \\ (a+b)(a^2-ab+b^2-a^2-2ab-b^2)=0 \\ -3ab(a+b)=0 \\ \Rightarrow a=0, b=0, a=-b \\ $The first two cases gives us $x=2$ and $x=\frac{1}{2} $ respectively$ \\ $ By inspection (can also be done through factoring the quadratic), the third case yields $x=1$

please do tell me if there's a much simpler and elegant way

Paradoxica · Mar 6, 2017

jathu123 said:
A straight forward method to find the real solutions:

$(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3 \\ $ Let $a\equiv 2^x-4, $ and $b\equiv 4^x-2 \\ \therefore a^3+b^3=(a+b)^3 \\ (a+b)(a^2-ab+b^2)=(a+b)^3 \\ (a+b)(a^2-ab+b^2-a^2-2ab-b^2)=0 \\ -3ab(a+b)=0 \\ \Rightarrow a=0, b=0, a=-b \\ $The first two cases gives us $x=2$ and $x=\frac{1}{2} $ respectively$ \\ $ By inspection (can also be done through factoring the quadratic), the third case yields $x=1$

please do tell me if there's a much simpler and elegant way

∄ a simpler way

Mahan1 · Mar 6, 2017

jathu123 said:
A straight forward method to find the real solutions:

$(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3 \\ $ Let $a\equiv 2^x-4, $ and $b\equiv 4^x-2 \\ \therefore a^3+b^3=(a+b)^3 \\ (a+b)(a^2-ab+b^2)=(a+b)^3 \\ (a+b)(a^2-ab+b^2-a^2-2ab-b^2)=0 \\ -3ab(a+b)=0 \\ \Rightarrow a=0, b=0, a=-b \\ $The first two cases gives us $x=2$ and $x=\frac{1}{2} $ respectively$ \\ $ By inspection (can also be done through factoring the quadratic), the third case yields $x=1$

please do tell me if there's a much simpler and elegant way

Your method is quite elegant and smart, but there is a slightly simpler method which uses this identity:

$(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac) =a^{3}+b^{3}+c^{3}-3abc$

$\textrm{Let} a= 2^{x} -4, b =4^{x}-2, c= 6-4^{x}-2^{x}$

hence a+b+c= 0, note $a^{3}+b^{3}+c^{3} = 0$

$\therefore a^{3}+b^{3}+c^{3} = 0 = 3abc \Rightarrow$

$a=0 \Rightarrow x=2$

$b =0 \Rightarrow x=\frac{1}{2}$

$c= 0 \Rightarrow 6 =4^{x}+2^{x} \Rightarrow y^{2}+y -6 =0 \ \textrm{where} \ y = 2^{x}$

$\textrm{hence} \ y = -3,2 \Rightarrow x =1$

Using this method, we get x=1 straightaway.

omegadot · Mar 8, 2017

Mahan1 said:
Your method is quite elegant and smart, but there is a slightly simpler method which uses this identity:

$(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac) =a^{3}+b^{3}+c^{3}-3abc$

$\textrm{Let} a= 2^{x} -4, b =4^{x}-2, c= 6-4^{x}-2^{x}$

hence a+b+c= 0, note $a^{3}+b^{3}+c^{3} = 0$

$\therefore a^{3}+b^{3}+c^{3} = 0 = 3abc \Rightarrow$

$a=0 \Rightarrow x=2$

$b =0 \Rightarrow x=\frac{1}{2}$

$c= 0 \Rightarrow 6 =4^{x}+2^{x} \Rightarrow y^{2}+y -6 =0 \ \textrm{where} \ y = 2^{x}$

$\textrm{hence} \ y = -3,2 \Rightarrow x =1$

Using this method, we get x=1 straightaway.

$$\noindent Not that I would wish this on anybody, but brute force also gives all three real solutions. Let $a = 2^x > 0$ for all $x \in \mathbb{R}.$

$\noindent Writing the equation as$

$(a - 4)^3 + (a^2 - 2)^3 - (a^2 + a - 6)^3 = 0$

$\noindent on factorising this becomes$

$-3(a - 2)(a + 3)(a - 4)(a^2 - 2) = 0.$

$\noindent Solving gives$

$\begin{align*} a = 2 &\Rightarrow 2^x = 2^1 \Rightarrow x = 1\\a = 4 &\Rightarrow 2^x = 2^2 \Rightarrow x = 2\\a = \sqrt{2} &\Rightarrow 2^x = 2^{1/2} \Rightarrow x = \frac{1}{2}.\end{align*}$

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Exponential equations (1 Viewer)

Mahan1

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Paradoxica

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