How do you solve for x when the natural log of the square root of x equals the square root of the natural log of x?

Answer 1

ln(√x)=√(lnx)

because √x = x^(1/2), ln(x^(1/2))=√(lnx)

using a logarithmic property, we can say that .5(lnx)=√(lnx)

now, pretend that lnx=y

.5y=√y

square both sides

.25y^2=y

subtract y from both sides

.25y^2 -y=0

factor

y(.25y - 1)=0

so either y=0 or .25y -1 =0

If .25y -1=0, then y=4

so lnx=0 or lnx=4

lnx cannot equal zero because lnx=0 means e^x=0 and that is impossible.

Now, we are left with lnx=4

Isolate x by making both sides of the equation powers of e:

e^(lnx)=e^4

x=e^4, which is approximately 54.6

Lastly, check this answer by plugging e^4 back into the original equation:

ln(√(e^4))=√(ln(e^4))

ln(e^2)=√(4(lne))

2lne=2√1

2(1)=2

2=2

There you go!