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Question Number 172111 by Mikenice last updated on 23/Jun/22

solve  2^x =4x

$${solve} \\ $$$$\mathrm{2}^{{x}} =\mathrm{4}{x} \\ $$

Commented by mr W last updated on 23/Jun/22

you have asked 2^x =10x. it is solved  generally for a^x =bx. so please don′t   post similar questions like  2^x =4x,   2^x =5x etc...

$${you}\:{have}\:{asked}\:\mathrm{2}^{{x}} =\mathrm{10}{x}.\:{it}\:{is}\:{solved} \\ $$$${generally}\:{for}\:{a}^{{x}} ={bx}.\:{so}\:{please}\:{don}'{t}\: \\ $$$${post}\:{similar}\:{questions}\:{like} \\ $$$$\mathrm{2}^{{x}} =\mathrm{4}{x},\: \\ $$$$\mathrm{2}^{{x}} =\mathrm{5}{x}\:{etc}... \\ $$

Answered by Mikenice last updated on 23/Jun/22

let 2^x =p  ⇒x=(√p)  p=4(√p)  take square to both side  p^2 =16p  p=16, then  2^x =16  2^x =2^4   x=4.

$${let}\:\mathrm{2}^{{x}} ={p} \\ $$$$\Rightarrow{x}=\sqrt{{p}} \\ $$$${p}=\mathrm{4}\sqrt{{p}} \\ $$$${take}\:{square}\:{to}\:{both}\:{side} \\ $$$${p}^{\mathrm{2}} =\mathrm{16}{p} \\ $$$${p}=\mathrm{16},\:{then} \\ $$$$\mathrm{2}^{{x}} =\mathrm{16} \\ $$$$\mathrm{2}^{{x}} =\mathrm{2}^{\mathrm{4}} \\ $$$${x}=\mathrm{4}. \\ $$$$ \\ $$$$ \\ $$

Commented by mr W last updated on 23/Jun/22

x=0.3099 is solution too.

$${x}=\mathrm{0}.\mathrm{3099}\:{is}\:{solution}\:{too}. \\ $$

Commented by mr W last updated on 23/Jun/22

please check following step sir:  let 2^x =p  ⇒x=(√p)  should it not be x=log_2  p ?

$${please}\:{check}\:{following}\:{step}\:{sir}: \\ $$$${let}\:\mathrm{2}^{{x}} ={p} \\ $$$$\Rightarrow{x}=\sqrt{{p}}\:\:{should}\:{it}\:{not}\:{be}\:{x}=\mathrm{log}_{\mathrm{2}} \:{p}\:? \\ $$

Commented by Mikenice last updated on 23/Jun/22

no, that is what it will be but you can try it by using in   your own way

$${no},\:{that}\:{is}\:{what}\:{it}\:{will}\:{be}\:{but}\:{you}\:{can}\:{try}\:{it}\:{by}\:{using}\:{in}\: \\ $$$${your}\:{own}\:{way} \\ $$

Commented by mr W last updated on 24/Jun/22

2^x =p ⇒x=(√p) is wrong.   not i say that, but the mathematics   says that.   IT IS NOT PERSONAL sir.

$$\mathrm{2}^{{x}} ={p}\:\Rightarrow{x}=\sqrt{{p}}\:{is}\:{wrong}.\: \\ $$$${not}\:{i}\:{say}\:{that},\:{but}\:{the}\:{mathematics}\: \\ $$$${says}\:{that}.\: \\ $$$${IT}\:{IS}\:{NOT}\:{PERSONAL}\:{sir}. \\ $$

Answered by Mikenice last updated on 24/Jun/22

let 2^x =y−−−1  x=log_2 y−−−2  equate eqn 2 by eqn1  y=4log_2 y,then divide both side by 4  (y/4)=log_2 y⇒y=2^(y/4) −−−3  sub eqn 3 into eqn 1  2^x =2^(y/4)   x=(y/4)−−−4  let y=16,then  x=((16)/4)⇒x=4.

$${let}\:\mathrm{2}^{{x}} ={y}−−−\mathrm{1} \\ $$$${x}={log}_{\mathrm{2}} {y}−−−\mathrm{2} \\ $$$${equate}\:{eqn}\:\mathrm{2}\:{by}\:{eqn}\mathrm{1} \\ $$$${y}=\mathrm{4}{log}_{\mathrm{2}} {y},{then}\:{divide}\:{both}\:{side}\:{by}\:\mathrm{4} \\ $$$$\frac{{y}}{\mathrm{4}}={log}_{\mathrm{2}} {y}\Rightarrow{y}=\mathrm{2}^{\frac{{y}}{\mathrm{4}}} −−−\mathrm{3} \\ $$$${sub}\:{eqn}\:\mathrm{3}\:{into}\:{eqn}\:\mathrm{1} \\ $$$$\mathrm{2}^{{x}} =\mathrm{2}^{\frac{{y}}{\mathrm{4}}} \\ $$$${x}=\frac{{y}}{\mathrm{4}}−−−\mathrm{4} \\ $$$${let}\:{y}=\mathrm{16},{then} \\ $$$${x}=\frac{\mathrm{16}}{\mathrm{4}}\Rightarrow{x}=\mathrm{4}. \\ $$

Answered by mr W last updated on 25/Jun/22

2^x =4x  e^(xln 2) =4x  (−xln 2)e^(−xln 2) =−((ln 2)/4)  −xln 2=W(−((ln 2)/4))  ⇒x=−((W(−((ln 2)/4)))/(ln 2))          = { ((−((−2.77258872)/(ln 2))=4)),((−((−0.21481112)/(ln 2))=0.309907)) :}    generally for  a^x =bx  ⇒x=−((W(−((ln a)/b)))/(ln a))

$$\mathrm{2}^{{x}} =\mathrm{4}{x} \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{4}{x} \\ $$$$\left(−{x}\mathrm{ln}\:\mathrm{2}\right){e}^{−{x}\mathrm{ln}\:\mathrm{2}} =−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}} \\ $$$$−{x}\mathrm{ln}\:\mathrm{2}={W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}\right) \\ $$$$\Rightarrow{x}=−\frac{{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}\right)}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\begin{cases}{−\frac{−\mathrm{2}.\mathrm{77258872}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{4}}\\{−\frac{−\mathrm{0}.\mathrm{21481112}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{0}.\mathrm{309907}}\end{cases} \\ $$$$ \\ $$$${generally}\:{for} \\ $$$$\boldsymbol{{a}}^{\boldsymbol{{x}}} =\boldsymbol{{bx}} \\ $$$$\Rightarrow\boldsymbol{{x}}=−\frac{\boldsymbol{{W}}\left(−\frac{\boldsymbol{\mathrm{ln}}\:\boldsymbol{{a}}}{\boldsymbol{{b}}}\right)}{\boldsymbol{\mathrm{ln}}\:\boldsymbol{{a}}} \\ $$

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