Question-68234

Question Number 68234 by Mikael last updated on 07/Sep/19

Commented by kaivan.ahmadi last updated on 07/Sep/19

3^(x+(1/2)) +3^(x−(1/2)) =4^x +2^(2x−1) ⇒ 3^(x−(1/2)) (3+1)=2^(2x−1) (2+1)⇒ (2^(2x−1) /3^(x−(1/2)) )=(4/3)⇒(2^(2x−1) /4)=(3^(x−(1/2)) /3)⇒2^(2x−3) =3^(x+(1/2)) ⇒ { ((2x−3=0⇒x=(3/2))),((x+(1/2)=0⇒x=−(1/2))) :} x=(3/2) is answer

$$\mathrm{3}^{{x}+\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{4}^{{x}} +\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} \Rightarrow \\ $$$$\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{3}+\mathrm{1}\right)=\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} \left(\mathrm{2}+\mathrm{1}\right)\Rightarrow \\ $$$$\frac{\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} }{\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{4}}{\mathrm{3}}\Rightarrow\frac{\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} }{\mathrm{4}}=\frac{\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{3}}\Rightarrow\mathrm{2}^{\mathrm{2}{x}−\mathrm{3}} =\mathrm{3}^{{x}+\frac{\mathrm{1}}{\mathrm{2}}} \Rightarrow \\ $$$$\begin{cases}{\mathrm{2}{x}−\mathrm{3}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}}}\\{{x}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\Rightarrow{x}=−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}\:{is}\:{answer} \\ $$

Commented by Mikael last updated on 07/Sep/19

$${thanks} \\ $$

Commented by Prithwish sen last updated on 08/Sep/19

2^(2x) =a 3^x =b a−(b/( (√3))) = (√3)b−(a/2) ((3a)/2)= ((4b)/( (√3))) (a/b) = (2^3 /3^(3/2) ) ∴ 2x=3⇒x=(3/2)

$$\mathrm{2}^{\mathrm{2x}} =\mathrm{a}\:\:\:\mathrm{3}^{\mathrm{x}} =\mathrm{b} \\ $$$$\mathrm{a}−\frac{\mathrm{b}}{\:\sqrt{\mathrm{3}}}\:=\:\sqrt{\mathrm{3}}\mathrm{b}−\frac{\mathrm{a}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3a}}{\mathrm{2}}=\:\frac{\mathrm{4b}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{a}}{\mathrm{b}}\:=\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\therefore\:\mathrm{2x}=\mathrm{3}\Rightarrow\mathrm{x}=\frac{\mathrm{3}}{\mathrm{2}}\: \\ $$

Commented by Rasheed.Sindhi last updated on 09/Sep/19

Sir kaivan ahmadi I don′t understand 2^(2x−3) =3^(x+(1/2)) ⇒^(?) { ((2x−3=0⇒x=(3/2))),((x+(1/2)=0⇒x=−(1/2))) :}

$$\mathrm{Sir}\:\mathrm{kaivan}\:\mathrm{ahmadi}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand} \\ $$$$\mathrm{2}^{\mathrm{2}{x}−\mathrm{3}} =\mathrm{3}^{{x}+\frac{\mathrm{1}}{\mathrm{2}}} \overset{?} {\Rightarrow}\begin{cases}{\mathrm{2}{x}−\mathrm{3}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}}}\\{{x}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\Rightarrow{x}=−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$

Commented by kaivan.ahmadi last updated on 13/Sep/19

since 2^0 =3^0 otherwise this equality is not true.

$${since}\:\mathrm{2}^{\mathrm{0}} =\mathrm{3}^{\mathrm{0}} \\ $$$${otherwise}\:{this}\:{equality}\:{is}\:{not}\:{true}. \\ $$

Answered by Rasheed.Sindhi last updated on 09/Sep/19

4^x +4^(x−(1/2)) =3^(x+(1/2)) +3^(x−(1/2)) 4^(x−(1/2)) (4^(1/2) +1)=3^(x−(1/2)) (3+1) 4^(x−(1/2)−1) =3^(x−(1/2)−1) (4^(x−(3/2)) /3^(x−(3/2)) )=1 ((4/3))^(x−(3/2)) =((4/3))^0 x−(3/2)=0 x=(3/2)

$$\mathrm{4}^{\mathrm{x}} +\mathrm{4}^{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{3}^{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}^{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{4}^{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}\right)=\mathrm{3}^{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{3}+\mathrm{1}\right) \\ $$$$\mathrm{4}^{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} =\mathrm{3}^{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \\ $$$$\frac{\mathrm{4}^{\mathrm{x}−\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}^{\mathrm{x}−\frac{\mathrm{3}}{\mathrm{2}}} }=\mathrm{1} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{x}−\frac{\mathrm{3}}{\mathrm{2}}} =\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{0}} \\ $$$$\mathrm{x}−\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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