化简 ln(ar(ax)^{r-1}e^{-(ax)^r})

解答

化简

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

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

求解步骤

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

使用对数计算法则:

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}})

使用对数计算法则:

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}})

使用对数计算法则:

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}

s im pl i f y 3 ln (2) + 4 ln (5)

e x p an d lo g_{10} (\frac{x ^{5}}{z y ^{3}})

s im pl i f y ln (e^{- t} sin (t))

s im pl i f y 2 \cdot ln (7) - 6 \cdot ln (z - 9)

s im pl i f y \frac{63}{17} ln ∣ - 4 ∣ - \frac{63}{17} ln ∣ - 8 ∣