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Question Number 147223 by mathlove last updated on 19/Jul/21

Commented by bramlexs22 last updated on 19/Jul/21

2^(3x) +2^x .3^(2x) −2.3^(3x) = 0 { ((2^x =u)),((3^x =v)) :} ⇒u^3 +u.v^2 −2v^3 = 0 ⇒1+((v/u))^2 −2((v/u))^3 = 0 ⇒2t^3 −t^2 −1=0 ⇒(t−1)(2t^2 +t+1)=0 for t=1 ⇒u=v ⇒2^x = 3^x ; ((2/3))^x =1 ; x=0 for 2t^2 +t+1=0 ; not real for x

$$\:\:\:\mathrm{2}^{\mathrm{3x}} +\mathrm{2}^{\mathrm{x}} .\mathrm{3}^{\mathrm{2x}} −\mathrm{2}.\mathrm{3}^{\mathrm{3x}} \:=\:\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{2}^{\mathrm{x}} =\mathrm{u}}\\{\mathrm{3}^{\mathrm{x}} =\mathrm{v}}\end{cases}\:\Rightarrow\mathrm{u}^{\mathrm{3}} +\mathrm{u}.\mathrm{v}^{\mathrm{2}} −\mathrm{2v}^{\mathrm{3}} =\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}+\left(\frac{\mathrm{v}}{\mathrm{u}}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{v}}{\mathrm{u}}\right)^{\mathrm{3}} =\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{2t}^{\mathrm{3}} −\mathrm{t}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{2t}^{\mathrm{2}} +\mathrm{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{t}=\mathrm{1}\:\Rightarrow\mathrm{u}=\mathrm{v}\: \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{3}^{\mathrm{x}} \:;\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} =\mathrm{1}\:;\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{2t}^{\mathrm{2}} +\mathrm{t}+\mathrm{1}=\mathrm{0}\:;\:\mathrm{not}\:\mathrm{real}\:\mathrm{for}\:\mathrm{x} \\ $$$$ \\ $$

Answered by puissant last updated on 19/Jul/21

⇒ 8^x +18^x =2×27^x ⇒((8/(18)))^x +1=2(((27)/(18)))^x ⇒((2/3))^(2x) +1=2((3/2))^x let A=((2/3))^x ⇒ A^(−1) =((3/2))^x ⇒A^2 +1=2A^(−1) ⇒A^3 +A−2=0 ⇒(A−1)(A^2 +A+2)=0 ⇒A=1 or A^2 +A+2=0(impossible) A=1 ⇒ ((2/3))^x =1 ⇒ e^(xln((2/3))) =1 ⇒xln((2/3))=0 ⇒ x=0.. S_R ={0}...

$$\Rightarrow\:\mathrm{8}^{\mathrm{x}} +\mathrm{18}^{\mathrm{x}} =\mathrm{2}×\mathrm{27}^{\mathrm{x}} \\ $$$$\Rightarrow\left(\frac{\mathrm{8}}{\mathrm{18}}\right)^{\mathrm{x}} +\mathrm{1}=\mathrm{2}\left(\frac{\mathrm{27}}{\mathrm{18}}\right)^{\mathrm{x}} \\ $$$$\Rightarrow\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2x}} +\mathrm{1}=\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} \: \\ $$$$\mathrm{let}\:\mathrm{A}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \Rightarrow\:\mathrm{A}^{−\mathrm{1}} =\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} \\ $$$$\Rightarrow\mathrm{A}^{\mathrm{2}} +\mathrm{1}=\mathrm{2A}^{−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{A}^{\mathrm{3}} +\mathrm{A}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{A}−\mathrm{1}\right)\left(\mathrm{A}^{\mathrm{2}} +\mathrm{A}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{A}=\mathrm{1}\:\mathrm{or}\:\mathrm{A}^{\mathrm{2}} +\mathrm{A}+\mathrm{2}=\mathrm{0}\left(\mathrm{impossible}\right) \\ $$$$\mathrm{A}=\mathrm{1}\:\Rightarrow\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} =\mathrm{1}\:\Rightarrow\:\mathrm{e}^{\mathrm{xln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{xln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{0}\:\Rightarrow\:\mathrm{x}=\mathrm{0}.. \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}_{\mathbb{R}} =\left\{\mathrm{0}\right\}... \\ $$

Commented by Rasheed.Sindhi last updated on 19/Jul/21

((8/(18)))^x =^(?) ((2/3))^(2x)

$$\left(\frac{\mathrm{8}}{\mathrm{18}}\right)^{\mathrm{x}} \overset{?} {=}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2x}} \\ $$

Commented by puissant last updated on 19/Jul/21

((8/(18)))^x =((4/9))^x =((2^2 /3^2 ))^x =(((2/3))^2 )^x =((2/3))^(2x) ..

$$\left(\frac{\mathrm{8}}{\mathrm{18}}\right)^{\mathrm{x}} =\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{\mathrm{x}} =\left(\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} }\right)^{\mathrm{x}} =\left(\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \right)^{\mathrm{x}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2x}} .. \\ $$

Commented by Rasheed.Sindhi last updated on 19/Jul/21

Ok sir, thank you

$$\mathrm{Ok}\:\mathrm{sir},\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by otchereabdullai@gmail.com last updated on 19/Jul/21

nice!

$$\mathrm{nice}! \\ $$

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