What is the answer to 2ln(x)+(ln(x))^2=0 ?

The answer to 2ln(x)+(ln(x))^2=0 is x=1,x= 1/(e^2)

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2ln(x)+(ln(x))^2=0

2 ln (x) + (ln (x))^{2} = 0

x = 1, x = \frac{1}{e ^{2}}

Decimal Notation

x = 1, x = 0.13533 \dots

Solution steps

2 ln (x) + (ln (x))^{2} = 0

Rewrite the equation with

ln (x) = u

2 u + (u)^{2} = 0

Solve

2 u + u^{2} = 0 : u = 0, u = - 2

u = 0, u = - 2

Substitute back

u = ln (x),

solve for

x

Solve

ln (x) = 0 : x = 1

Solve

ln (x) = - 2 : x = \frac{1}{e ^{2}}

x = 1, x = \frac{1}{e ^{2}}

Verify Solutions:

x = 1

True

, x = \frac{1}{e ^{2}}

True

The solutions are

x = 1, x = \frac{1}{e ^{2}}

lo g_{x} (\frac{1}{125}) = - 3

lo g_{5} (x - 4) = 2

ln (\frac{25}{25 - y}) = ln (y)

lo g_{2} (x - 1) = 6 - lo g_{2} (3 x + 1)

0.61 = ln (y)

What is the answer to 2ln(x)+(ln(x))^2=0 ?
The answer to 2ln(x)+(ln(x))^2=0 is x=1,x= 1/(e^2)
Solve for x: 2ln(x)+(ln(x))^2=0
The solution is x=1,x= 1/(e^2)