What is ln^2(1/e)+(-1/2)-ln^2(-1) ?

The solution to ln^2(1/e)+(-1/2)-ln^2(-1) is -ln^2(-1)-3/2

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simplify ln^2(1/e)+(-1/2)-ln^2(-1)

simplify

ln^{2} (\frac{1}{e}) + (- \frac{1}{2}) - ln^{2} (- 1)

- ln^{2} (- 1) - \frac{3}{2}

Solution steps

ln^{2} (\frac{1}{e}) + (- \frac{1}{2}) - ln^{2} (- 1)

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

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

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

ln^{2} (1) = 0

ln^{2} (e) = 1

= 0 - 1 - \frac{1}{2} - ln^{2} (- 1)

Subtract the numbers:

0 - 1 = - 1

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

Combine the fractions

- 1 - \frac{1}{2} : - \frac{3}{2}

= - ln^{2} (- 1) - \frac{3}{2}

lo g_{10} (y^{7} x z^{3})

3 ln (x) - \frac{1}{2} ln (3 x + 1) + 5 ln (x - 1) - ln (x + 5)

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

3 lo g_{10} (a) - lo g_{10} (q)

What is ln^2(1/e)+(-1/2)-ln^2(-1) ?
The solution to ln^2(1/e)+(-1/2)-ln^2(-1) is -ln^2(-1)-3/2