If-x-2-2-x-find-x-please-i-need-workings-

Question Number 6216 by sanusihammed last updated on 18/Jun/16

$${If}\:\:\:\:\:{x}^{\mathrm{2}} \:\:=\:\:\mathrm{2}^{{x}} \:\:\:\:\:\:{find}\:\:{x}\: \\ $$$$ \\ $$$${please}\:{i}\:{need}\:{workings} \\ $$

Answered by prakash jain last updated on 18/Jun/16

2ln x=xln 2 ((ln 2)/2)=((ln x)/x) .....(A) (1/2)ln 2=((ln x)/e^(ln x) ) (1/2)ln 2=(ln x)e^(−ln x) −(1/2)ln 2=(−ln x)e^(−ln x) W(−(1/2)ln 2)=−ln x .....(1) calculating W(−(1/2)ln 2) (ln (1/2))e^(ln (1/2)) =(1/2)ln (1/2)=(1/2)(−ln 2)=−(1/2)ln 2 ⇒W(−(1/2)ln 2)=ln (1/2) ....(2) substituting W(−(1/2)ln 2)=ln (1/2) from (2) to (1) ln (1/2)=−ln x⇒x=2

$$\mathrm{2ln}\:{x}={x}\mathrm{ln}\:\mathrm{2} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{ln}\:{x}}{{x}}\:\:\:\:\:\:\:…..\left({A}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}=\frac{\mathrm{ln}\:{x}}{{e}^{\mathrm{ln}\:{x}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}=\left(\mathrm{ln}\:{x}\right){e}^{−\mathrm{ln}\:{x}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}=\left(−\mathrm{ln}\:{x}\right){e}^{−\mathrm{ln}\:{x}} \\ $$$${W}\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}\right)=−\mathrm{ln}\:{x}\:\:\:\:\:…..\left(\mathrm{1}\right) \\ $$$${calculating}\:{W}\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\left(\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}\right){e}^{\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{ln}\:\mathrm{2}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$\Rightarrow{W}\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}\right)=\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:….\left(\mathrm{2}\right) \\ $$$$\mathrm{substituting}\:{W}\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}\right)=\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{from}\:\left(\mathrm{2}\right)\:\mathrm{to}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}=−\mathrm{ln}\:{x}\Rightarrow{x}=\mathrm{2} \\ $$

Commented by sanusihammed last updated on 18/Jun/16

$${Thanks}. \\ $$

Commented by FilupSmith last updated on 19/Jun/16

Wow I am having trouble understanding the last half, good work!

$$\mathrm{Wow}\:\mathrm{I}\:\mathrm{am}\:\mathrm{having}\:\mathrm{trouble}\:\mathrm{understanding} \\ $$$$\mathrm{the}\:\mathrm{last}\:\mathrm{half},\:\mathrm{good}\:\mathrm{work}! \\ $$

Commented by malwaan last updated on 19/Jun/16

$${can}\:{you}\:{prove}\:{the}\:\mathrm{2}{nd}\:{case}\: \\ $$$${x}=\mathrm{4} \\ $$

Commented by prakash jain last updated on 19/Jun/16

ln (1/4)e^(ln (1/4)) =(1/4)ln (1/4)=(1/2)ln (1/2)=−(1/2)ln 2 W(−(1/2)ln 2) is multivalued W(−(1/2)ln 2)=ln (1/2) or ln (1/4) x=2 or 4

$$\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{4}}{e}^{\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{4}}} =\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$${W}\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}\right)\:\mathrm{is}\:\mathrm{multivalued} \\ $$$${W}\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}\right)=\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{or}\:\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}=\mathrm{2}\:{or}\:\mathrm{4} \\ $$

Answered by malwaan last updated on 19/Jun/16

we have two cases case 1: x=2 ⇒2^2 =2^2 =4 case 2 : x=4 ⇒2^4 =4^2 =16

$${we}\:{have}\:{two}\:{cases} \\ $$$${case}\:\mathrm{1}:\:\:{x}=\mathrm{2}\:\Rightarrow\mathrm{2}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${case}\:\mathrm{2}\::\:\:{x}=\mathrm{4}\:\Rightarrow\mathrm{2}^{\mathrm{4}} =\mathrm{4}^{\mathrm{2}} =\mathrm{16} \\ $$

Facebook Tweet Pin

If-x-2-2-x-find-x-please-i-need-workings-

Leave a Reply Cancel reply