What is ln(ar(ax)^{r-1}e^{-(ax)^r}) ?

The solution to ln(ar(ax)^{r-1}e^{-(ax)^r}) is ln(a)+ln(r)+ln(ax)(r-1)-(ax)^r

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simplify ln(ar(ax)^{r-1}e^{-(ax)^r})

simplify

ln (a r (a x)^{r - 1} e^{- (a x)^{r}})

ln (a) + ln (r) + ln (a x) (r - 1) - (a x)^{r}

Solution steps

ln (a r (a x)^{r - 1} e^{- (a x)^{r}})

Apply log rule:

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

ln (a r (a x)^{r - 1} e^{- (a x)^{r}}) = ln (a) + ln (r) + ln ((a x)^{r - 1}) + ln (e^{- (a x)^{r}})

= ln (a) + ln (r) + ln ((a x)^{r - 1}) + ln (e^{- (a x)^{r}})

Apply log rule:

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

ln ((a x)^{r - 1}) = (r - 1) ln (a x)

= ln (a) + ln (r) + (r - 1) ln (a x) + ln (e^{- (a x)^{r}})

Apply log rule:

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

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

= ln (a) + ln (r) + (r - 1) ln (a x) - (a x)^{r} ln (e)

(a x)^{r} ln (e) = (a x)^{r}

= ln (a) + ln (r) + ln (a x) (r - 1) - (a x)^{r}

3 ln (2) + 4 ln (5)

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

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

2 \cdot ln (7) - 6 \cdot ln (z - 9)

What is ln(ar(ax)^{r-1}e^{-(ax)^r}) ?
The solution to ln(ar(ax)^{r-1}e^{-(ax)^r}) is ln(a)+ln(r)+ln(ax)(r-1)-(ax)^r