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Solve for x.
x^2 = 2^x
Do I take the natural log on both sides as step 1?
x^2 = 2^x
Do I take the natural log on both sides as step 1?
x^2 = 2^x
prepare for Lambert form xe^(-(ln(2)x)/2)=1
rewrite the equation with -(ln(2)x)/2)=u and x=-(2u)/ln(2)
-(2u)/ln(2)*e^u=1
rewrite -(2u)/ln(2)*e^u=1 in Lambert form (e^u)u=-ln(2)/2
solve e^u=-ln(2)/2=> u=-2ln(2), u=-1*ln(2)
substitute back u=-(ln(2)x)/2), solve for x
-2ln(2)=-(ln(2)x)/2
-4ln(2)=-ln(2)x
-4=-x
x=4
-1*ln(2)=-(ln(2)x)/2
-2*ln(2)=-ln(2)x
-2=-x
x=2
Lambert's function, also called the "W" function (Lambert W-Function -- from Wolfram MathWorld) is defined as the inverse function to f(x)= xe^x.
Did you try looking for him yourself?
Johann Heinrich Lambert (1728 - 1777) - Biography - MacTutor History of Mathematics (st-andrews.ac.uk)