Menu Close

2-x-3-x-6-x-9-x-find-x-

Tinku Tara 0 Comments
Pin




Question Number 200330 by hardmath last updated on 17/Nov/23
2^x − 3^x = (√(6^x − 9^x )) find: x = ?
$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:=\:\sqrt{\mathrm{6}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{9}^{\boldsymbol{\mathrm{x}}} } \\ $$$$\mathrm{find}:\:\:\:\mathrm{x}\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 17/Nov/23
(2^x −3^x )^2 =6^x −9^x 2^(2x) −2(2^x )(3^x )+3^(2x) =6^x −9^x 4^x −2(6^x )+9^x =6^x −9^x 4^x +2(9^x )=3(6^x ) ((4/6))^x +2((9/6))^x =3 ((2/3))^x +2((3/2))^x =3 let ((2/3))^x =a a+(2/a)=3 a^2 −3a+2=0 (a−1)(a−2)=0 a=1,2 •((2/3))^x =1=((2/3))^0 ⇒x=0✓ •((2/3))^x =2 xlog_2 ((2/3))=log_2 2=1 x=(1/(log_2 2 −log_2 3 )) =(1/(1−log_2 3 ))✓
$$\left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} \right)^{\mathrm{2}} =\mathrm{6}^{{x}} −\mathrm{9}^{{x}} \\ $$$$\mathrm{2}^{\mathrm{2}{x}} −\mathrm{2}\left(\mathrm{2}^{{x}} \right)\left(\mathrm{3}^{{x}} \right)+\mathrm{3}^{\mathrm{2}{x}} =\mathrm{6}^{{x}} −\mathrm{9}^{{x}} \\ $$$$\mathrm{4}^{{x}} −\mathrm{2}\left(\mathrm{6}^{{x}} \right)+\mathrm{9}^{{x}} =\mathrm{6}^{{x}} −\mathrm{9}^{{x}} \\ $$$$\mathrm{4}^{{x}} +\mathrm{2}\left(\mathrm{9}^{{x}} \right)=\mathrm{3}\left(\mathrm{6}^{{x}} \right) \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{6}}\right)^{{x}} +\mathrm{2}\left(\frac{\mathrm{9}}{\mathrm{6}}\right)^{{x}} =\mathrm{3} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} +\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} =\mathrm{3} \\ $$$${let}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} ={a} \\ $$$${a}+\frac{\mathrm{2}}{{a}}=\mathrm{3} \\ $$$${a}^{\mathrm{2}} −\mathrm{3}{a}+\mathrm{2}=\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)\left({a}−\mathrm{2}\right)=\mathrm{0} \\ $$$${a}=\mathrm{1},\mathrm{2} \\ $$$$\bullet\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{1}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{0}} \Rightarrow{x}=\mathrm{0}\checkmark \\ $$$$\bullet\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{2} \\ $$$$\:\:\:{x}\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{log}_{\mathrm{2}} \mathrm{2}=\mathrm{1} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \mathrm{2}\:−\mathrm{log}_{\mathrm{2}} \mathrm{3}\:}\:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{log}_{\mathrm{2}} \mathrm{3}\:}\checkmark \\ $$
Commented by Rasheed.Sindhi last updated on 18/Nov/23
I always remember an old forum member ′dinusar′,′hongking′; when I see your posts
$${I}\:{always}\:{remember}\:{an}\:{old}\:{forum} \\ $$$${member}\:'{dinusar}','{hongking}';\: \\ $$$$\:{when}\:{I}\:{see}\:{your}\:{posts} \\ $$
Commented by hardmath last updated on 18/Nov/23
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Nov/23
2^x − 3^x = (√(6^x − 9^x )) 2^x − 3^x =3^(x/2) (√(2^x − 3^x )) 2^x − 3^x =3^(x/2) (√(2^x − 3^x )) ((√(2^x − 3^x )) )^2 −3^(x/2) (√(2^x − 3^x )) (√(2^x − 3^x )) ((√(2^x − 3^x )) −3^(x/2) )=0 (√(2^x − 3^x )) =0 ∨ (√(2^x − 3^x )) =3^(x/2) ▶2^x =3^x ⇒((2/3))^x =1=((2/3))^0 ⇒x=0 ▶(√(2^x − 3^x )) =3^(x/2) 2^x − 3^x =3^x 2^x =2∙3^x ((2/3))^x =2 log_2 ((2/3))^x =log_2 2 xlog_2 ((2/3))=1 x=(1/(log_2 2−log_2 3 ))=(1/(1−log_2 3))
$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:=\:\sqrt{\mathrm{6}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{9}^{\boldsymbol{\mathrm{x}}} } \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:=\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{2}}} \:\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} } \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:=\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{2}}} \:\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} } \\ $$$$\left(\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} }\:\right)^{\mathrm{2}} −\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{2}}} \:\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} } \\ $$$$\:\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} }\:\left(\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} }\:−\mathrm{3}^{{x}/\mathrm{2}} \right)=\mathrm{0} \\ $$$$\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} }\:=\mathrm{0}\:\vee\:\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} }\:=\mathrm{3}^{{x}/\mathrm{2}} \\ $$$$\blacktriangleright\mathrm{2}^{{x}} =\mathrm{3}^{{x}} \Rightarrow\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{1}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{0}} \Rightarrow{x}=\mathrm{0} \\ $$$$\blacktriangleright\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} }\:=\mathrm{3}^{\mathrm{x}/\mathrm{2}} \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:=\mathrm{3}^{\mathrm{x}} \\ $$$$\mathrm{2}^{\mathrm{x}} =\mathrm{2}\centerdot\mathrm{3}^{\mathrm{x}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} =\mathrm{2} \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} =\mathrm{log}_{\mathrm{2}} \mathrm{2} \\ $$$$\mathrm{xlog}_{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{1} \\ $$$$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \mathrm{2}−\mathrm{log}_{\mathrm{2}} \mathrm{3}\:\:}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{log}_{\mathrm{2}} \mathrm{3}} \\ $$
Answered by witcher3 last updated on 17/Nov/23
(√(1(1−((3/2))^x ))=((2/3))^(x/2) −((3/2))^(x/2) (√(1−a^2 ))=((1/a))−a a=1;x=>p0 ;or a=(√(1−a^2 )) ⇔a=(1/( (√2)))⇒((3/2))^(x/2) =(1/( (√2))) ⇔x=2((ln((1/( (√2)))))/(ln((3/2))))
$$\sqrt{\mathrm{1}\left(\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} \right.}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\frac{\mathrm{x}}{\mathrm{2}}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{x}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }=\left(\frac{\mathrm{1}}{\mathrm{a}}\right)−\mathrm{a} \\ $$$$\mathrm{a}=\mathrm{1};\mathrm{x}=>\mathrm{p0}\:;\mathrm{or}\:\mathrm{a}=\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\mathrm{a}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\Rightarrow\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\frac{\mathrm{x}}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Leftrightarrow\mathrm{x}=\mathrm{2}\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)}{\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$
Answered by manxsol last updated on 18/Nov/23
$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *