Solve #x^sqrtx = 16# ? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Cesareo R. · NickTheTurtle Apr 6, 2017 #x=4# Explanation: Making #y = sqrt(x)# we have #(y^2)^y=y^(2y)=4^2# so #y^y=4# so #y = 2# and #x = y^2= 4# Now using the Lambert function #W# we have #y = log(4)/(W(log(4)))# and also #x = y^2=(log(4)/(W(log(4))))^2 = 4# Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 1208 views around the world You can reuse this answer Creative Commons License