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

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

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1-ln^2(x)=0

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

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

Solution steps

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

Rewrite the equation with

ln (x) = u

1 - u^{2} = 0

Solve

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

u = 1, u = - 1

Substitute back

u = ln (x),

solve for

x

Solve

ln (x) = 1 : x = e

Solve

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

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

Verify Solutions:

x = e

True

, x = \frac{1}{e}

True

The solutions are

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

lo g_{4} (\frac{1}{4} x + 2) = 1

x, y = ln (x^{2})

c = 2 (ln (x) + x)

y, ln (y) = - ln (x) + C

N, lo g_{10} (N) = 0.3 + 2 t