What is ln(αβx^{β-1})(e^{-αx^β}) ?

The solution to ln(αβx^{β-1})(e^{-αx^β}) is (ln(x)(β-1)+ln(αβ))/(e^{αx^β)}

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simplify ln(αβx^{β-1})(e^{-αx^β})

simplify

ln (α β x^{β - 1}) (e^{- α x^{β}})

\frac{ln ( x ) ( β - 1 ) + ln ( α β )}{e ^{α x^{β}}}

Solution steps

ln (α β x^{β - 1}) (e^{- α x^{β}})

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{α x^{β}}} ln (α β x^{β - 1})

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ln ( α β x ^{β - 1} )}{e ^{α x^{β}}}

Multiply:

1 \cdot ln (α β x^{β - 1}) = ln (α β x^{β - 1})

= \frac{ln ( α β x ^{β - 1} )}{e ^{α x^{β}}}

Simplify

ln (α β x^{β - 1}) : ln (x) (β - 1) + ln (α β)

= \frac{ln ( α β ) + ln ( x ) ( β - 1 )}{e ^{α x^{β}}}

= \frac{ln ( x ) ( β - 1 ) + ln ( α β )}{e ^{α x^{β}}}

9 (1.05)^{- (- 8)}

- 1.602 x 1 0^{- 19}

(j - 1)^{2} (j + 1)^{2}

8.35 \cdot 1 0^{- 4}

e^{- s t} (sin (t))

What is ln(αβx^{β-1})(e^{-αx^β}) ?
The solution to ln(αβx^{β-1})(e^{-αx^β}) is (ln(x)(β-1)+ln(αβ))/(e^{αx^β)}