What is ln(θe^{-n(x^1+x^2+x^3)}) ?

The solution to ln(θe^{-n(x^1+x^2+x^3)}) is ln(θ)-n(x+x^2+x^3)

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simplify ln(θe^{-n(x^1+x^2+x^3)})

simplify

ln (θ e^{- n (x^{1} + x^{2} + x^{3})})

ln (θ) - n (x + x^{2} + x^{3})

Solution steps

ln (θ e^{- n (x^{1} + x^{2} + x^{3})})

Apply log rule:

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

ln (θ e^{- n (x^{1} + x^{2} + x^{3})}) = ln (θ) + ln (e^{- n (x^{1} + x^{2} + x^{3})})

= ln (θ) + ln (e^{- n (x^{3} + x^{2} + x^{1})})

Apply log rule:

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

ln (e^{- n (x^{1} + x^{2} + x^{3})}) = - n (x^{1} + x^{2} + x^{3}) ln (e)

= ln (θ) - n (x^{1} + x^{2} + x^{3}) ln (e)

n (x^{1} + x^{2} + x^{3}) ln (e) = n (x + x^{2} + x^{3})

= ln (θ) - n (x^{3} + x^{2} + x)

ln (\frac{225 - 0.9}{( 2.82108 - 0.9 )})

- \frac{1}{3} ln (\frac{5 + 1}{5 - 2})

ln (\frac{x ^{3}}{y ^{2}})

lo g_{0.5} (x^{2})

4 ln (\frac{97}{100})

What is ln(θe^{-n(x^1+x^2+x^3)}) ?
The solution to ln(θe^{-n(x^1+x^2+x^3)}) is ln(θ)-n(x+x^2+x^3)