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

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

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

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

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

Decimal Notation

x = 1, x = 0.13533 \dots

Solution steps

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

Rewrite the equation with

ln (x) = u

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

Solve

u^{2} + 2 u = 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_{4} (1 + x^{2}) = 0.25

ln (x) = 1 + ln (x + 2)

2 = - ln (x)

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

0.65 = lo g_{10} (x)

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