What is the answer to (ln(x))^2-ln(x^2)+1=0 ?

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

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(ln(x))^2-ln(x^2)+1=0

(ln (x))^{2} - ln (x^{2}) + 1 = 0

x = e

Decimal

x = 2.71828 \dots

Solution steps

(ln (x))^{2} - ln (x^{2}) + 1 = 0

Apply log rules

(ln (x))^{2} - 2 ln (x) + 1 = 0

Rewrite the equation with

ln (x) = u

(u)^{2} - 2 u + 1 = 0

Solve

u^{2} - 2 u + 1 = 0 : u = 1

u = 1

Substitute back

u = ln (x),

solve for

x

Solve

ln (x) = 1 : x = e

x = e

Verify Solutions:

x = e

True

The solution is

x = e

lo g_{2} (3 - x) = 0

ln (4 x e^{x - 1}) = x

lo g_{x} (0.125) = 2

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

1 + lo g_{x} (125) = 6

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