What is ln((e^{-10θ}θ^{20})/(207360)) ?

The solution to ln((e^{-10θ}θ^{20})/(207360)) is -10θ+20ln(θ)-ln(207360)

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simplify ln((e^{-10θ}θ^{20})/(207360))

simplify

ln (\frac{e ^{- 10 θ} θ ^{20}}{207360})

- 10 θ + 20 ln (θ) - ln (207360)

Solution steps

ln (\frac{e ^{- 10 θ} θ ^{20}}{207360})

Apply log rule:

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

ln (\frac{e ^{- 10 θ} θ ^{20}}{207360}) = ln (e^{- 10 θ} θ^{20}) - ln (207360)

= ln (e^{- 10 θ} θ^{20}) - ln (207360)

Apply log rule:

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

ln (e^{- 10 θ} θ^{20}) = ln (e^{- 10 θ}) + ln (θ^{20})

= ln (e^{- 10 θ}) + ln (θ^{20}) - ln (207360)

Apply log rule:

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

ln (e^{- 10 θ}) = - 10 θ ln (e)

= - 10 θ ln (e) + ln (θ^{20}) - ln (207360)

Apply log rule:

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

ln (θ^{20}) = 20 ln (θ)

= - 10 θ ln (e) + 20 ln (θ) - ln (207360)

10 θ ln (e) = 10 θ

= - 10 θ + 20 ln (θ) - ln (207360)

lo g_{10} (y \cdot z)

lo g_{10} (\frac{1}{200})

lo g_{10} (\frac{x - 5}{x + 4})

ln (3 ln (e^{e}))

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

What is ln((e^{-10θ}θ^{20})/(207360)) ?
The solution to ln((e^{-10θ}θ^{20})/(207360)) is -10θ+20ln(θ)-ln(207360)