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

The solution to ln((e^{-c}(1+c))/(c^2)) is -c+ln(1+c)-2ln(c)

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simplify ln((e^{-c}(1+c))/(c^2))

simplify

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

- c + ln (1 + c) - 2 ln (c)

Solution steps

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

Apply log rule:

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

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

= ln (e^{- c} (c + 1)) - ln (c^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (c^{2}) = 2 ln (c)

= ln (e^{- c} (1 + c)) - 2 ln (c)

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (e^{- c} (1 + c)) = ln (e^{- c}) + ln (1 + c)

= ln (e^{- c}) + ln (c + 1) - 2 ln (c)

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (e^{- c}) = - c ln (e)

= - c ln (e) + ln (1 + c) - 2 ln (c)

c ln (e) = c

= - c + ln (c + 1) - 2 ln (c)

lo g_{10} (\frac{z ^{5} y ^{4}}{x ^{5}})

lo g_{2} (x) + 8 lo g_{2} (y)

- lo g_{10} (2.5 \cdot 1 0^{- 11})

\frac{1}{2} ln (4) - 4 ln (5 d)

ln (\frac{e \cdot x}{5 y ^{4}})

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