#x^2+x(4^x-14)+4^(x+1)-72=0# The value of x?

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Answer 1 · 2017-08-05T11:01:08

#x = {-4,2}#

#4^x(x+4)+x^2-14x-72=4^x(x+4)+(x+4)(x-18)=0#

This is equivalent to

#{(4^x+x-18=0),(x+4=0):}#

Now #4^x+x-18=0# can be solved using the Lambert function #W(cdot)#

because calling #e^lambda = 4# we have

#e^(lambda x)+x-18=0# now making #y = x-18# we have

#e^(lambda (y+18))+y=0# or

#e^(18lambda)e^(lambda y)+y=0# or

#e^(18lambda)=-y e^(-lambda y)# or

#lambda e^(18lambda)=(-lambda y)e^(-lambda y)#

Now using the property

#Xe^X = Y hArr X = W(Y)#

we have

#-lambda y = W(lambda e^(18lambda))# or

#y = -1/lambda W(lambda e^(18lambda)) = x-18# and then

#x = 18-1/lambda W(lambda e^(18lambda)) =2#

Here #lambda = log_e 4#

and finally the solution set is

#x = {-4,2}#