Solve for x

Solve for x.

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

 

What is the Lambert form? This is beyond precalculus, right?

 

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.

 

Who's Lambert?

 

Did you try looking for him yourself?
Johann Heinrich Lambert (1728 - 1777) - Biography - MacTutor History of Mathematics (st-andrews.ac.uk)

 

I could have done a search for him.