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Question Number 17742 by tawa tawa last updated on 10/Jul/17

x^3 = 3^x , find x.

$$\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{3}^{\mathrm{x}} ,\:\:\:\mathrm{find}\:\mathrm{x}. \\ $$

Commented by 1kanika# last updated on 10/Jul/17

x=3

$$\mathrm{x}=\mathrm{3} \\ $$

Answered by mrW1 last updated on 10/Jul/17

x=3^(x/3) =e^((x/3)ln 3) xe^(−(x/3)ln 3) =1 (−(x/3)ln 3)e^(−(x/3)ln 3) =−((ln 3)/3) −(x/3)ln 3=W(−((ln 3)/3)) ⇒x=−(3/(ln 3))W(−((ln 3)/3))= { ((2.4...)),(3) :}

$$\mathrm{x}=\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{3}}} =\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}} \\ $$$$\mathrm{xe}^{−\frac{\mathrm{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}} =\mathrm{1} \\ $$$$\left(−\frac{\mathrm{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}\right)\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}} =−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}} \\ $$$$−\frac{\mathrm{x}}{\mathrm{3}}\mathrm{ln}\:\mathrm{3}=\mathrm{W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{x}=−\frac{\mathrm{3}}{\mathrm{ln}\:\mathrm{3}}\mathrm{W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}}\right)=\begin{cases}{\mathrm{2}.\mathrm{4}...}\\{\mathrm{3}}\end{cases} \\ $$

Commented by tawa tawa last updated on 10/Jul/17

God bless you sir. i understand it. i really appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{understand}\:\mathrm{it}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by mrW1 last updated on 10/Jul/17

solve x^2 =3^x

$$\mathrm{solve}\:\mathrm{x}^{\mathrm{2}} =\mathrm{3}^{\mathrm{x}} \\ $$

Commented by tawa tawa last updated on 10/Jul/17

x^2 = 3^x 3^(x/2) = x e^((x/2)ln3 ) = x 1 = xe^(−((xln3)/2)) −((ln3)/2) = (−((xln3)/2))e^((−((xln3)/2))) −((xln3)/2) = We^((−((ln3)/2))) x = ((−2We^((−((ln3)/2))) )/(ln3)) x = ((−2We^((((ln3)/2))) )/(ln3)) x = ((− 2 × 0.376839)/(1.09861)) x = ((− 0.753674)/(1.09861)) x = − 0.6860

$$\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{3}^{\mathrm{x}} \\ $$$$\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{2}}} \:=\:\mathrm{x} \\ $$$$\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{2}}\mathrm{ln3}\:} =\:\mathrm{x} \\ $$$$\mathrm{1}\:=\:\mathrm{xe}^{−\frac{\mathrm{xln3}}{\mathrm{2}}} \\ $$$$−\frac{\mathrm{ln3}}{\mathrm{2}}\:=\:\left(−\frac{\mathrm{xln3}}{\mathrm{2}}\right)\mathrm{e}^{\left(−\frac{\mathrm{xln3}}{\mathrm{2}}\right)} \\ $$$$−\frac{\mathrm{xln3}}{\mathrm{2}}\:=\:\mathrm{We}^{\left(−\frac{\mathrm{ln3}}{\mathrm{2}}\right)} \\ $$$$\mathrm{x}\:=\:\frac{−\mathrm{2We}^{\left(−\frac{\mathrm{ln3}}{\mathrm{2}}\right)} }{\mathrm{ln3}} \\ $$$$\mathrm{x}\:=\:\frac{−\mathrm{2We}^{\left(\frac{\mathrm{ln3}}{\mathrm{2}}\right)} }{\mathrm{ln3}} \\ $$$$\mathrm{x}\:=\:\frac{−\:\mathrm{2}\:×\:\mathrm{0}.\mathrm{376839}}{\mathrm{1}.\mathrm{09861}} \\ $$$$\mathrm{x}\:=\:\frac{−\:\mathrm{0}.\mathrm{753674}}{\mathrm{1}.\mathrm{09861}} \\ $$$$\mathrm{x}\:=\:−\:\mathrm{0}.\mathrm{6860} \\ $$

Commented by mrW1 last updated on 10/Jul/17

x^2 = 3^x ⇒±3^(x/2) = x ...... totally 3 solutions

$$\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{3}^{\mathrm{x}} \\ $$$$\Rightarrow\pm\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{2}}} \:=\:\mathrm{x} \\ $$$$...... \\ $$$$\mathrm{totally}\:\mathrm{3}\:\mathrm{solutions} \\ $$

Commented by tawa tawa last updated on 10/Jul/17

ohh, i see.... it almost affect the last answer

$$\mathrm{ohh},\:\mathrm{i}\:\mathrm{see}....\:\mathrm{it}\:\mathrm{almost}\:\mathrm{affect}\:\mathrm{the}\:\mathrm{last}\:\mathrm{answer} \\ $$

Commented by tawa tawa last updated on 10/Jul/17

Sir.. .. x^3 has two solutions x^2 has three solutions ... or it depends on the question.

$$\mathrm{Sir}..\:..\:\mathrm{x}^{\mathrm{3}} \:\mathrm{has}\:\mathrm{two}\:\mathrm{solutions} \\ $$$$\mathrm{x}^{\mathrm{2}} \:\mathrm{has}\:\mathrm{three}\:\mathrm{solutions}\:... \\ $$$$\mathrm{or}\:\mathrm{it}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{the}\:\mathrm{question}. \\ $$

Commented by tawa tawa last updated on 10/Jul/17

Show me sir.

$$\mathrm{Show}\:\mathrm{me}\:\mathrm{sir}. \\ $$

Commented by mrW1 last updated on 10/Jul/17

if x^3 =3^x since 3^x >0 ⇒x^3 >0 ⇒x>0 ⇒x^3 =3^x ≡x=3^(x/3) if x^2 =3^x x can also be negative, ⇒x^2 =3^x ≡x=±3^(x/2) just like x^3 =8 ⇒x=(8)^(1/3) =2 x^3 =−8 ⇒x=(−8)^(1/3) =−2 but x^2 =4 ⇒x=±(4)^(1/2) =±2

$$\mathrm{if}\:\mathrm{x}^{\mathrm{3}} =\mathrm{3}^{\mathrm{x}} \\ $$$$\mathrm{since}\:\mathrm{3}^{\mathrm{x}} >\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{3}} >\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{3}} =\mathrm{3}^{\mathrm{x}} \equiv\mathrm{x}=\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{3}}} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{x}^{\mathrm{2}} =\mathrm{3}^{\mathrm{x}} \\ $$$$\mathrm{x}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\:\mathrm{negative}, \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} =\mathrm{3}^{\mathrm{x}} \equiv\mathrm{x}=\pm\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{2}}} \\ $$$$ \\ $$$$\mathrm{just}\:\mathrm{like} \\ $$$$\mathrm{x}^{\mathrm{3}} =\mathrm{8} \\ $$$$\Rightarrow\mathrm{x}=\left(\mathrm{8}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{2} \\ $$$$\mathrm{x}^{\mathrm{3}} =−\mathrm{8} \\ $$$$\Rightarrow\mathrm{x}=\left(−\mathrm{8}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =−\mathrm{2} \\ $$$$\mathrm{but} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\mathrm{x}=\pm\left(\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\pm\mathrm{2} \\ $$

Commented by tawa tawa last updated on 10/Jul/17

Now, i understand sir. God bless you

$$\mathrm{Now},\:\mathrm{i}\:\mathrm{understand}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

Commented by mrW1 last updated on 10/Jul/17

solution for x^4 =3^x x=±3^(x/4) x=±e^((x/4)ln 3) xe^(−(x/4)ln 3) =±1 (−(x/4)ln 3)e^(−(x/4)ln 3) =±((ln 3)/4) −(x/4)ln 3=W(±((ln 3)/4)) ⇒x=−(4/(ln 3))W(±((ln 3)/4)) x= { ((−(4/(ln 3)) W(((ln 3)/4))=−0.80225)),((−(4/(ln 3)) W(−((ln 3)/4))= { ((1.51687)),((7.17476)) :})) :}

$$\mathrm{solution}\:\mathrm{for}\:\mathrm{x}^{\mathrm{4}} =\mathrm{3}^{\mathrm{x}} \\ $$$$\mathrm{x}=\pm\mathrm{3}^{\frac{\mathrm{x}}{\mathrm{4}}} \\ $$$$\mathrm{x}=\pm\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{4}}\mathrm{ln}\:\mathrm{3}} \\ $$$$\mathrm{xe}^{−\frac{\mathrm{x}}{\mathrm{4}}\mathrm{ln}\:\mathrm{3}} =\pm\mathrm{1} \\ $$$$\left(−\frac{\mathrm{x}}{\mathrm{4}}\mathrm{ln}\:\mathrm{3}\right)\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{4}}\mathrm{ln}\:\mathrm{3}} =\pm\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4}} \\ $$$$−\frac{\mathrm{x}}{\mathrm{4}}\mathrm{ln}\:\mathrm{3}=\mathrm{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\mathrm{x}=−\frac{\mathrm{4}}{\mathrm{ln}\:\mathrm{3}}\mathrm{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4}}\right) \\ $$$$\mathrm{x}=\begin{cases}{−\frac{\mathrm{4}}{\mathrm{ln}\:\mathrm{3}}\:\mathrm{W}\left(\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4}}\right)=−\mathrm{0}.\mathrm{80225}}\\{−\frac{\mathrm{4}}{\mathrm{ln}\:\mathrm{3}}\:\mathrm{W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4}}\right)=\begin{cases}{\mathrm{1}.\mathrm{51687}}\\{\mathrm{7}.\mathrm{17476}}\end{cases}}\end{cases} \\ $$

Commented by tawa tawa last updated on 10/Jul/17

I really appreciate your effort sir. God bless you sir.

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

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