Solve for x

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
 
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?
 

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