What is ln(1/2 y^4x^{11}e^{-yx^3}) ?

The solution to ln(1/2 y^4x^{11}e^{-yx^3}) is -yx^3+4ln(y)+11ln(x)-ln(2)

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simplify ln(1/2 y^4x^{11}e^{-yx^3})

simplify

ln (\frac{1}{2} y^{4} x^{11} e^{- y x^{3}})

- y x^{3} + 4 ln (y) + 11 ln (x) - ln (2)

Solution steps

ln (\frac{1}{2} y^{4} x^{11} e^{- y x^{3}})

Apply log rule:

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

ln (\frac{1}{2} y^{4} x^{11} e^{- y x^{3}}) = ln (\frac{1}{2}) + ln (y^{4}) + ln (x^{11}) + ln (e^{- y x^{3}})

= ln (\frac{1}{2}) + ln (y^{4}) + ln (x^{11}) + ln (e^{- y x^{3}})

Apply log rule:

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

ln (y^{4}) = 4 ln (y)

= ln (\frac{1}{2}) + 4 ln (y) + ln (x^{11}) + ln (e^{- y x^{3}})

Apply log rule:

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

ln (x^{11}) = 11 ln (x)

= ln (\frac{1}{2}) + 4 ln (y) + 11 ln (x) + ln (e^{- y x^{3}})

Apply log rule:

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

ln (e^{- y x^{3}}) = - y x^{3} ln (e)

= ln (\frac{1}{2}) + 4 ln (y) + 11 ln (x) - y x^{3} ln (e)

Apply log rule:

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

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

= ln (1) - ln (2) + 4 ln (y) + 11 ln (x) - ln (e) y x^{3}

ln (1) = 0

y x^{3} ln (e) = y x^{3}

= 0 - ln (2) + 4 ln (y) + 11 ln (x) - y x^{3}

0 - ln (2) + 4 ln (y) + 11 ln (x) - y x^{3} = - ln (2) + 4 ln (y) + 11 ln (x) - y x^{3}

= - y x^{3} + 4 ln (y) + 11 ln (x) - ln (2)

lo g_{4} (5) - lo g_{4} (3)

\frac{lo g _{10} ( 2 x ) + lo g _{10} ( 4 x ^{2} )}{3}

4 ln (θ) + 9 ln (1 - θ)

2 lo g_{b} (2) - lo g_{b} (7) - 4 lo g_{b} (5)

lo g_{10} (30) - lo g_{10} (5)

What is ln(1/2 y^4x^{11}e^{-yx^3}) ?
The solution to ln(1/2 y^4x^{11}e^{-yx^3}) is -yx^3+4ln(y)+11ln(x)-ln(2)