What is ln((t+1)((4t^3)/3)(2t^2)(C)) ?

The solution to ln((t+1)((4t^3)/3)(2t^2)(C)) is ln(t+1)++ln(4)+3ln(t)-ln(3)+ln(2)+2ln(t)+ln(C)

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simplify ln((t+1)((4t^3)/3)(2t^2)(C))

simplify

ln ((t + 1) (\frac{4 t ^{3}}{3}) (2 t^{2}) (C))

ln (t + 1) + + ln (4) + 3 ln (t) - ln (3) + ln (2) + 2 ln (t) + ln (C)

Solution steps

ln ((t + 1) (\frac{4 t ^{3}}{3}) (2 t^{2}) (C))

Apply log rule:

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

ln ((t + 1) (\frac{4 t ^{3}}{3}) (2 t^{2}) (C)) = ln (t + 1) + ln (\frac{4 t ^{3}}{3}) + ln (2 t^{2}) + ln (C)

= ln (t + 1) + ln (\frac{4 t ^{3}}{3}) + ln (2 t^{2}) + ln (C)

Apply log rule:

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

ln (\frac{4 t ^{3}}{3}) = ln (4 t^{3}) - ln (3)

= ln (t + 1) + + ln (4 t^{3}) - ln (3) + ln (2 t^{2}) + ln (C)

Apply log rule:

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

ln (2 t^{2}) = ln (2) + ln (t^{2})

= ln (t + 1) + + ln (4 t^{3}) - ln (3) + ln (2) + ln (t^{2}) + ln (C)

Apply log rule:

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

ln (t^{2}) = 2 ln (t)

= ln (t + 1) + + ln (4 t^{3}) - ln (3) + ln (2) + 2 ln (t) + ln (C)

Apply log rule:

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

ln (4 t^{3}) = ln (4) + ln (t^{3})

= ln (t + 1) + + ln (4) + ln (t^{3}) - ln (3) + ln (2) + 2 ln (t) + ln (C)

Apply log rule:

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

ln (t^{3}) = 3 ln (t)

= ln (t + 1) + + ln (4) + 3 ln (t) - ln (3) + ln (2) + 2 ln (t) + ln (C)

ln (3 \cdot 1 0^{- 3})

ln (- e^{- 4 x}) + ln (e^{4 x})

15 ln (15)^{2} + 7 - 15 ln (0)^{2} + 7

4 ln (c) (ln (a) - 3 ln (b))

\frac{1}{2} ln (x) + \frac{1}{2} ln (\frac{p x}{p y} x)

What is ln((t+1)((4t^3)/3)(2t^2)(C)) ?
The solution to ln((t+1)((4t^3)/3)(2t^2)(C)) is ln(t+1)++ln(4)+3ln(t)-ln(3)+ln(2)+2ln(t)+ln(C)