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

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

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

simplify

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

- ln (- 1)

Solution steps

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

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

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

Apply log rule:

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

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

Apply rule

\frac{a}{1} = a

= - ln (- 1)

231.84357 - 224.62838 - 8.314 ln (\frac{1}{5.2})

2 lo g_{10} (x) - lo g_{10} (y)

lo g_{2} (\frac{s ^{7}}{5 t ^{2}})

ln (\frac{m ^{3}}{n ^{7}})

lo g_{10} (\frac{9 a ^{16}}{b ^{2}})

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