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Question Number 114551 by bemath last updated on 19/Sep/20

The value of x in the given equation 4^x −3^(x−(1/2)) = 3^(x+(1/2)) −2^(2x−1) , is _,

$${The}\:{value}\:{of}\:{x}\:{in}\:{the}\:{given}\:{equation} \\ $$$$\mathrm{4}^{{x}} −\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{3}^{{x}+\frac{\mathrm{1}}{\mathrm{2}}} \:−\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} \:,\:{is}\:\_, \\ $$

Commented by Dwaipayan Shikari last updated on 19/Sep/20

2^(2x) +2^(2x−1) =3^(x+(1/2)) +3^(x−(1/2)) 2^(2x) (1+(1/2))=3^(x−(1/2)) (3+1) 2^(2x−1) .3=3^(x−(1/2)) .2^2 2^(2x−3) =3^((2x−3)/2) (2x−3)log2=((2x−3)/2)log(3) (2x−3)(log2−(1/2)log3)=0 2x−3=0 x=(3/2)

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

Commented by Rasheed.Sindhi last updated on 19/Sep/20

Alternate way(to prevent log) From above: 2^(2x−3) =3^((2x−3)/2) ((2/( (√3))))^(2x−3) =((2/( (√3))))^0 2x−3=0 x=(3/2)

$${Alternate}\:{way}\left({to}\:{prevent}\:{log}\right)\: \\ $$$${From}\:{above}: \\ $$$$\mathrm{2}^{\mathrm{2}{x}−\mathrm{3}} =\mathrm{3}^{\frac{\mathrm{2}{x}−\mathrm{3}}{\mathrm{2}}} \\ $$$$\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}{x}−\mathrm{3}} =\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{0}} \\ $$$$\mathrm{2}{x}−\mathrm{3}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by Dwaipayan Shikari last updated on 19/Sep/20

Yes sir!

$${Yes}\:{sir}! \\ $$

Answered by bobhans last updated on 19/Sep/20

⇔ 4^x +(1/2).4^x = ((√3)+(1/( (√3)))).3^x ⇔ (3/2).4^x = (4/3)(√3).3^x ⇔ 2^(2x−3) = 3^(x−(3/2)) ⇒(2x−3)ln (2)=(x−(3/2))ln (3) ⇒x(2 ln 2−ln 3)=3 ln 2−(3/2)ln 3 ⇒x = (1/2)(((6ln 2−3ln 3)/(2ln 2−ln 3))) = (3/2)(((2ln 2−ln 3)/(2ln 2−ln 3))) ⇒x = (3/2)

$$\:\Leftrightarrow\:\mathrm{4}^{{x}} +\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{4}^{{x}} \:=\:\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right).\mathrm{3}^{{x}} \\ $$$$\Leftrightarrow\:\frac{\mathrm{3}}{\mathrm{2}}.\mathrm{4}^{{x}} \:=\:\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\mathrm{3}}.\mathrm{3}^{{x}} \\ $$$$\Leftrightarrow\:\mathrm{2}^{\mathrm{2}{x}−\mathrm{3}} \:=\:\mathrm{3}^{{x}−\frac{\mathrm{3}}{\mathrm{2}}} \:\Rightarrow\left(\mathrm{2}{x}−\mathrm{3}\right)\mathrm{ln}\:\left(\mathrm{2}\right)=\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)\mathrm{ln}\:\left(\mathrm{3}\right) \\ $$$$\Rightarrow{x}\left(\mathrm{2}\:\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}\right)=\mathrm{3}\:\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mathrm{3} \\ $$$$\Rightarrow{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{6ln}\:\mathrm{2}−\mathrm{3ln}\:\mathrm{3}}{\mathrm{2ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{2ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}}{\mathrm{2ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}}\right) \\ $$$$\Rightarrow{x}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by bemath last updated on 19/Sep/20

gave kudos

$${gave}\:{kudos}\: \\ $$

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