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Find-the-value-of-x-2-x-4x-workings-is-needed-please-




Question Number 5712 by sanusihammed last updated on 24/May/16
Find the value of x     2^x  = 4x    workings is needed please.
Findthevalueofx2x=4xworkingsisneededplease.
Commented by Yozzii last updated on 24/May/16
2^x =4x  ⇒2^(x−2) =x  Try x=4. ⇒2^(4−2) =4=2^2   2^2 =2^2 ∴ x=4 is one solution.  2^(x−2) >0 ∀x∈R⇒if 2^(x−2) =x⇒x>0 if  real solutions are to exist.  If x∈N⇒ x is even since x=2^(x−2)   and 2^(x−2)  is even for x−2≥1⇒x≥3.  For x≥6 2^(x−2) ≥2^(6−2) =16 and 2^(x−2)  diverges  rapidly (note that (d/dx)(2^(x−2) )=2^(x−2) ln2) whereas x rises at  a linear rate ((d/dx)(x)=1<2^(x−2) ln2 for x≥3).  This indicates that no further integer  solutions could possibly exist for 2^x =4x.    Graphing the two functions y=4x  and y=2^x    (x∈R) shows that some other  root lies in the interval [0,1].  Any method of determining the root  approximately could be used : Interval  bisection, Newton−Raphson method, Linear  interpolation, etc...
2x=4x2x2=xTryx=4.242=4=2222=22x=4isonesolution.2x2>0xRif2x2=xx>0ifrealsolutionsaretoexist.IfxNxisevensincex=2x2and2x2isevenforx21x3.Forx62x2262=16and2x2divergesrapidly(notethatddx(2x2)=2x2ln2)whereasxrisesatalinearrate(ddx(x)=1<2x2ln2forx3).Thisindicatesthatnofurtherintegersolutionscouldpossiblyexistfor2x=4x.Graphingthetwofunctionsy=4xandy=2x(xR)showsthatsomeotherrootliesintheinterval[0,1].Anymethodofdeterminingtherootapproximatelycouldbeused:Intervalbisection,NewtonRaphsonmethod,Linearinterpolation,etc
Commented by sanusihammed last updated on 24/May/16
Thaks. no algeraic solution.?
Thaks.noalgeraicsolution.?
Commented by Yozzii last updated on 25/May/16
My level of knowledge of math  cannot let me say that an algebraic  solution exists or not.
Mylevelofknowledgeofmathcannotletmesaythatanalgebraicsolutionexistsornot.

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