What is the answer to (ln(x))^2+4ln(x)+4=0 ?

The answer to (ln(x))^2+4ln(x)+4=0 is x= 1/(e^2)

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(ln(x))^2+4ln(x)+4=0

(ln (x))^{2} + 4 ln (x) + 4 = 0

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

Solution steps

(ln (x))^{2} + 4 ln (x) + 4 = 0

Rewrite the equation with

ln (x) = u

(u)^{2} + 4 u + 4 = 0

Solve

u^{2} + 4 u + 4 = 0 : u = - 2

u = - 2

Substitute back

u = ln (x),

solve for

x

Solve

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

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

Verify Solutions:

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

True

The solution is

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

lo g_{5} (x^{2}) = lo g_{5} (3 x + 4)

3 - ln (x) = \frac{x}{e} + 1

lo g_{10} (x) = 2.6

1 = 10 lo g_{10} (\frac{x}{1})

lo g_{4} ((2 - 3 x)^{2}) = 1

What is the answer to (ln(x))^2+4ln(x)+4=0 ?
The answer to (ln(x))^2+4ln(x)+4=0 is x= 1/(e^2)
Solve for x: (ln(x))^2+4ln(x)+4=0
The solution is x= 1/(e^2)