Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 58832 by otchereabdullai@gmail.com last updated on 30/Apr/19

$$\mathrm{find}\mathrm{x}\mathrm{if}{\mathrm{x}}^2={16}^{\mathrm{x}}$$

Commented by tanmay last updated on 01/May/19

$$f\left(x\right)=x^2-{16}^x$$$$f\left(0\right)<0$$$$f\left(1\right)<0$$$$sonorootbetween(0,1)$$$$f(-1)=1-\frac{1}{16}>0$$$$sorootliesbetween(-1,0)\leftarrow locationofroot$$$$f\left(x\right)=x^2-{\left(4^x\right)}^2=(x+4^x)(x-4^x)$$$$ontrialifweputx=-0.5$$$$f(-0.5)=(-0.5+4^{-0.5})(-0.5-4^{-0.5})$$$$=\{{(-0.5)}^2-{\left(4^{-0.5}\right)}^2\}$$$$=\{0.25-\left(4^{-1}\right)\}$$$$=0.25-0.25=0$$$$sox=-0.5isthesolution$$$$ihavelocatedthelocationofrootandfound$$$$bytrialmethod...\mathit{T}heremustbesomedirect$$$$methodtofindtheroot.$$$$$$

Commented by otchereabdullai@gmail.com last updated on 30/Apr/19

$$\mathrm{thanks}\mathrm{alot}\mathrm{prof}\mathrm{tanmay}$$

Commented by George Mark Samuel last updated on 30/Apr/19

$${\mathrm{x}}^2={16}^{\mathrm{x}}$$$${\mathrm{x}}^2=4^{\left(2\right)\mathrm{x}}$$$${\mathrm{x}}^2=2^{2\left(2\mathrm{x}\right)}$$$${\mathrm{x}}^2=2^{4\mathrm{x}}$$$$\mathrm{x}=2.\mid 2=4\mathrm{x}$$$$\mid \mathrm{divide}\mathrm{both}\mathrm{sides}\mathrm{by}4$$$$\mid \frac{2}{4}=\frac{4\mathrm{x}}{4}$$$$\mid \mathrm{x}=\frac{1}{2}$$$$\therefore \mathrm{x}=2\mathrm{or}\mathrm{x}=\frac{1}{2.}$$$$$$

Commented by MJS last updated on 01/May/19

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}$$$$x=2\Rightarrow x^2=4\wedge {16}^x=256\mathrm{but}4\ne 256$$$$x=\frac{1}{2}\Rightarrow x^2=\frac{1}{4}\wedge {16}^x=4\mathrm{but}\frac{1}{4}\ne 4$$

Answered by MJS last updated on 30/Apr/19

$$\mathrm{obviously}$$$$x⩾0\Rightarrow {16}^x>x^2$$$$x⩽-1\Rightarrow x^2>{16}^x$$$$\Rightarrow -1<x<0$$$$\mathrm{trying}\mathrm{we}\mathrm{find}$$$$x=-\frac{1}{2}\Rightarrow {(-\frac{1}{2})}^2={16}^{-\frac{1}{2}}=\frac{1}{4}$$

Commented by otchereabdullai@gmail.com last updated on 30/Apr/19

$$\mathrm{thanks}\mathrm{prof}$$

Answered by mr W last updated on 01/May/19

$$directmethodusingLambertfunction:$$$$x^2={16}^x$$$$\Rightarrow x=\pm {16}^{\frac{x}{2}}$$$$\Rightarrow x=\pm e^{\frac{x\ln 16}{2}}$$$$\Rightarrow {xe}^{-\frac{x\ln 16}{2}}=\pm 1$$$$\Rightarrow (-\frac{x\ln 16}{2})e^{-\frac{x\ln 16}{2}}=\mp \frac{\ln 16}{2}$$$$\Rightarrow -\frac{x\ln 16}{2}=\mathbb{W}(\mp \frac{\ln 16}{2})$$$$\Rightarrow x=-\frac{2}{\ln 16}\times \mathbb{W}(\mp \frac{\ln 16}{2})\leftarrow norealsolutionforW(-\frac{\ln 16}{2})!$$$$\Rightarrow x=-\frac{2\times 0.69314718}{\ln 16}=-0.5$$$$$$$$Bonus:$$$$ingeneral,if{x}^2=a^x$$$$\Rightarrow x=-\frac{2}{\ln a}\times \mathbb{W}(\pm \frac{\ln a}{2})$$$$e.g.a=2,i.e.x^2=2^x$$$$\Rightarrow x=-\frac{2}{\ln 2}\times \mathbb{W}(\pm \frac{\ln 2}{2})=\{\begin{array}{l}-\frac{2}{\ln 2}\times W\left(\frac{\ln 2}{2}\right)=-\frac{2}{\ln 2}\times 0.26570574=-0.766665\\ -\frac{2}{\ln 2}\times W(-\frac{\ln 2}{2})=\{\begin{array}{l}-\frac{2}{\ln 2}\times (-0.69314718)=2\\ -\frac{2}{\ln 2}\times (-1.38629436)=4\end{array}\end{array}$$

Commented by otchereabdullai@gmail.com last updated on 01/May/19

$$\mathrm{thank}\mathrm{you}\mathrm{soo}\mathrm{much}\mathrm{prof}\mathrm{W}$$

Commented by malwaan last updated on 02/May/19

$$HowcanIcalc.\mathbb{W}function?$$

Commented by mr W last updated on 02/May/19

$$ifyou{can}^{\prime }tfindapropercalculator,$$$$youcangetthefunctionvaluein$$$$thisway:$$$$theroot\left(s\right)ofeqn.$$$$x{e}^x=a$$$$is\left(are\right)thefunctionvalue\left(s\right)of\mathbb{W}\left(a\right).$$

Commented by malwaan last updated on 03/May/19

$$Iwilltry$$$$thankyousir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com