What is ln((m!e^{-x})/((1+x)^{m+1)}) ?

The solution to ln((m!e^{-x})/((1+x)^{m+1)}) is ln(m!)-x-ln(1+x)(m+1)

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simplify ln((m!e^{-x})/((1+x)^{m+1)})

simplify

ln (\frac{m ! e ^{- x}}{( 1 + x ) ^{m + 1}})

ln (m!) - x - ln (1 + x) (m + 1)

Solution steps

ln (\frac{m ! e ^{- x}}{( 1 + x ) ^{m + 1}})

Apply log rule:

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

ln (\frac{m ! e ^{- x}}{( 1 + x ) ^{m + 1}}) = ln (m! e^{- x}) - ln ((1 + x)^{m + 1})

= ln (e^{- x} m!) - ln ((x + 1)^{m + 1})

Apply log rule:

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

ln ((1 + x)^{m + 1}) = (m + 1) ln (1 + x)

= ln (m! e^{- x}) - (m + 1) ln (1 + x)

Apply log rule:

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

ln (m! e^{- x}) = ln (m!) + ln (e^{- x})

= ln (m!) + ln (e^{- x}) - ln (x + 1) (m + 1)

Apply log rule:

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

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

= ln (m!) - x ln (e) - (m + 1) ln (1 + x)

x ln (e) = x

= ln (m!) - x - ln (x + 1) (m + 1)

ln (x \cdot x)

lo g_{10} (\frac{[ 0.5 ]}{[ 0.4 ]}) + 3.77

ln (\frac{2 x + 1}{x + 2})

lo g_{e} (\frac{y}{2})

\frac{5 ln ( \frac{29}{74} )}{ln ( \frac{59}{74} )}

What is ln((m!e^{-x})/((1+x)^{m+1)}) ?
The solution to ln((m!e^{-x})/((1+x)^{m+1)}) is ln(m!)-x-ln(1+x)(m+1)