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

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

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

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

x = 1, x = e^{2}

Decimal Notation

x = 1, x = 7.38905 \dots

Solution steps

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

Apply log rules

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 = e^{2}

x = 1, x = e^{2}

Verify Solutions:

x = 1

True

, x = e^{2}

True

The solutions are

x = 1, x = e^{2}

lo g_{X} (\frac{2}{3}) = 4

lo g_{x} (3 - 1) = - 1

lo g_{3} (6 x + 34) = lo g_{3} (- 2)

lo g_{10} (4 x) + 13 = 14

lo g_{2} (lo g_{5} (x)) = 3

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