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Question Number 200696 by Bayat last updated on 22/Nov/23
Answered by aleks041103 last updated on 22/Nov/23
(2^x −3^x )^2 =4^x +9^x −2.6^x =6^x −9^x ⇒4^x +2.9^x −3.6^x =0 (√(4^x 9^x ))=6^x ⇒a+2b−3(√(ab))=0,a,b>0 ⇒(a+2b)^2 =9ab ⇒a^2 +4b^2 +4ab=9ab ⇒a^2 +4b^2 −5ab=0 ⇒4((b/a))^2 −5((b/a))+1=0 ⇒(b/a)=((9/4))^x =((5±(√(25−16)))/8)=((5±3)/8)=(1/4);1 ⇒x_(1,2) =log_(9/4) (1/4);0=((−ln(4))/(−ln(4/9)));0= =((ln(4))/(ln(4)−ln(9)));0=(1/(1−ln(9)/ln(4)));0= =(1/(1−ln(3)/ln(2)));0=(1/(1−log_2 (3)));0 ⇒ x_(1,2) =(1/(1−log_2 (3))) ; 0 = log_(9/4) (1/4) ; 0
$$\left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} \right)^{\mathrm{2}} =\mathrm{4}^{{x}} +\mathrm{9}^{{x}} −\mathrm{2}.\mathrm{6}^{{x}} =\mathrm{6}^{{x}} −\mathrm{9}^{{x}} \\ $$$$\Rightarrow\mathrm{4}^{{x}} +\mathrm{2}.\mathrm{9}^{{x}} −\mathrm{3}.\mathrm{6}^{{x}} =\mathrm{0} \\ $$$$\sqrt{\mathrm{4}^{{x}} \mathrm{9}^{{x}} }=\mathrm{6}^{{x}} \\ $$$$\Rightarrow{a}+\mathrm{2}{b}−\mathrm{3}\sqrt{{ab}}=\mathrm{0},{a},{b}>\mathrm{0} \\ $$$$\Rightarrow\left({a}+\mathrm{2}{b}\right)^{\mathrm{2}} =\mathrm{9}{ab} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{ab}=\mathrm{9}{ab} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} −\mathrm{5}{ab}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} −\mathrm{5}\left(\frac{{b}}{{a}}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{b}}{{a}}=\left(\frac{\mathrm{9}}{\mathrm{4}}\right)^{{x}} =\frac{\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{16}}}{\mathrm{8}}=\frac{\mathrm{5}\pm\mathrm{3}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}};\mathrm{1} \\ $$$$\Rightarrow{x}_{\mathrm{1},\mathrm{2}} ={log}_{\mathrm{9}/\mathrm{4}} \left(\mathrm{1}/\mathrm{4}\right);\mathrm{0}=\frac{−{ln}\left(\mathrm{4}\right)}{−{ln}\left(\mathrm{4}/\mathrm{9}\right)};\mathrm{0}= \\ $$$$=\frac{{ln}\left(\mathrm{4}\right)}{{ln}\left(\mathrm{4}\right)−{ln}\left(\mathrm{9}\right)};\mathrm{0}=\frac{\mathrm{1}}{\mathrm{1}−{ln}\left(\mathrm{9}\right)/{ln}\left(\mathrm{4}\right)};\mathrm{0}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{ln}\left(\mathrm{3}\right)/{ln}\left(\mathrm{2}\right)};\mathrm{0}=\frac{\mathrm{1}}{\mathrm{1}−{log}_{\mathrm{2}} \left(\mathrm{3}\right)};\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}−{log}_{\mathrm{2}} \left(\mathrm{3}\right)}\:;\:\mathrm{0}\:=\:{log}_{\mathrm{9}/\mathrm{4}} \left(\mathrm{1}/\mathrm{4}\right)\:;\:\mathrm{0} \\ $$
Answered by Rasheed.Sindhi last updated on 22/Nov/23
(2^x −3^x )^2 =6^x −9^x 4^x +9^x −2∙6^x −6^x +9^x =0 4^x +2(9^x )−3(6^x )=0 (4^x /6^x )+((2(9^x ))/6^x )−3=0 ((2/3))^x +2((3/2))^x −3=0 ((2/3))^x =y y+(2/y)−3=0 y^2 −3y+2=0 (y−1)(y−2)=0 y=1,2 ((2/3))^x =1,2 { ((((2/3))^x =((2/3))^0 ⇒x=0)),((((2/3))^x =2⇒2^(x−1) =3^x )) :} (x−1)log_2 2=xlog_2 3 x−xlog_2 3=1 x=(1/(1−log_2 3 ))
$$\left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} \right)^{\mathrm{2}} =\mathrm{6}^{{x}} −\mathrm{9}^{{x}} \\ $$$$\mathrm{4}^{{x}} +\mathrm{9}^{{x}} −\mathrm{2}\centerdot\mathrm{6}^{{x}} −\mathrm{6}^{{x}} +\mathrm{9}^{{x}} =\mathrm{0} \\ $$$$\mathrm{4}^{{x}} +\mathrm{2}\left(\mathrm{9}^{{x}} \right)−\mathrm{3}\left(\mathrm{6}^{{x}} \right)=\mathrm{0} \\ $$$$\frac{\mathrm{4}^{{x}} }{\mathrm{6}^{{x}} }+\frac{\mathrm{2}\left(\mathrm{9}^{{x}} \right)}{\mathrm{6}^{{x}} }−\mathrm{3}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} +\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} −\mathrm{3}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} ={y} \\ $$$${y}+\frac{\mathrm{2}}{{y}}−\mathrm{3}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} −\mathrm{3}{y}+\mathrm{2}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left({y}−\mathrm{2}\right)=\mathrm{0} \\ $$$${y}=\mathrm{1},\mathrm{2} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{1},\mathrm{2} \\ $$$$\begin{cases}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{0}} \Rightarrow{x}=\mathrm{0}}\\{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{2}\Rightarrow\mathrm{2}^{{x}−\mathrm{1}} =\mathrm{3}^{{x}} }\end{cases} \\ $$$$\left({x}−\mathrm{1}\right)\mathrm{log}_{\mathrm{2}} \mathrm{2}={x}\mathrm{log}_{\mathrm{2}} \mathrm{3} \\ $$$${x}−{x}\mathrm{log}_{\mathrm{2}} \mathrm{3}=\mathrm{1} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{log}_{\mathrm{2}} \mathrm{3}\:} \\ $$
Commented by Bayat last updated on 22/Nov/23
tnhaks alote
Commented by Bayat last updated on 22/Nov/23
Answered by Rasheed.Sindhi last updated on 22/Nov/23
(2^x −3^x )^2 =3^x (2^x −3^x ) (2^x −3^x )^2 −3^x (2^x −3^x )=0 (2^x −3^x )(2^x −3^x −3^x )=0 2^x =3^x ∨ 2^x =2(3^x ) x=0 ∨ 2^(x−1) =3^x (x−1)log_2 2=xlog_2 3 (x−1)(1)=xlog_2 3 x=(1/(1−log_2 3))
$$\:\:\left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} \right)^{\mathrm{2}} =\mathrm{3}^{{x}} \left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} \right) \\ $$$$\:\:\left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} \right)^{\mathrm{2}} −\mathrm{3}^{{x}} \left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} \right)=\mathrm{0} \\ $$$$\left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} \right)\left(\mathrm{2}^{{x}} −\mathrm{3}^{{x}} −\mathrm{3}^{{x}} \right)=\mathrm{0} \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{x}} \:\:\:\vee\:\:\mathrm{2}^{{x}} =\mathrm{2}\left(\mathrm{3}^{{x}} \right) \\ $$$$\:{x}=\mathrm{0}\:\vee\:\mathrm{2}^{{x}−\mathrm{1}} =\mathrm{3}^{{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\left({x}−\mathrm{1}\right)\mathrm{log}_{\mathrm{2}} \mathrm{2}={x}\mathrm{log}_{\mathrm{2}} \mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\left({x}−\mathrm{1}\right)\left(\mathrm{1}\right)={x}\mathrm{log}_{\mathrm{2}} \mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{log}_{\mathrm{2}} \mathrm{3}}\:\: \\ $$
Answered by witcher3 last updated on 22/Nov/23
6^x −9^x =(2^x −3^x ).3^x ⇔(2^x −3^x )(2^x −3^x )−(2^x −3^x ).3^x =0 ⇔(2^x −3^x )(2^x −.2.3^x )=0 2^x =3^x ⇒x=0 2^x −2.3^x =0 x=.((ln(2))/(ln(2)−ln(3)))
$$\mathrm{6}^{\mathrm{x}} −\mathrm{9}^{\mathrm{x}} =\left(\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\mathrm{x}} \right).\mathrm{3}^{\mathrm{x}} \\ $$$$\Leftrightarrow\left(\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\mathrm{x}} \right)\left(\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\mathrm{x}} \right)−\left(\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\mathrm{x}} \right).\mathrm{3}^{\mathrm{x}} =\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\mathrm{x}} \right)\left(\mathrm{2}^{\mathrm{x}} −.\mathrm{2}.\mathrm{3}^{\mathrm{x}} \right)=\mathrm{0} \\ $$$$\mathrm{2}^{\mathrm{x}} =\mathrm{3}^{\mathrm{x}} \Rightarrow\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{2}^{\mathrm{x}} −\mathrm{2}.\mathrm{3}^{\mathrm{x}} =\mathrm{0} \\ $$$$\mathrm{x}=.\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{3}\right)} \\ $$

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