What is the inverse of #y=log_4( x-3) +2x?# ?

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Answer 1 · 2018-02-26T09:09:35

#x = 1/2(6+W(2^2y-11))#

We can solve this problem using the so called Lambert function #W(cdot)#

#y = lnabs(x-3)/ln4 + 2x rArr y ln4=lnabs(x-3)+2x ln4#

Now making #z = x-3#

#e^(y ln4)=z e^(2(z+3)ln4) = ze^(2z)e^(6 ln4)# or

#e^((y-6)ln4)=z e^(2z)# or

#2 e^((y-6)ln4) = 2z e^(2z)#

Now using the equivalence

#Y = X e^X rArr X = W(Y)#

#2z=W(2 e^((y-6)ln4)) rArr z = 1/2 W(2 e^((y-6)ln4)) #

and finally

#x = 1/2 W(2 e^((y-6)ln4))+3# that can be simplified to

#x = 1/2(6+W(2^(2y-11)))#