What is 3ln(1+i)+ln((1+i)^5) ?

The solution to 3ln(1+i)+ln((1+i)^5) is 4ln(2)

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simplify 3ln(1+i)+ln((1+i)^5)

simplify

3 ln (1 + i) + ln ((1 + i)^{5})

4 ln (2)

Decimal Notation

2.77258 \dots

Solution steps

3 ln (1 + i) + ln ((1 + i)^{5})

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

3 ln (1 + i) = ln ((1 + i)^{3})

= ln ((1 + i)^{3}) + ln ((1 + i)^{5})

Apply log rule:

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

ln ((1 + i)^{3}) + ln ((1 + i)^{5}) = ln ((1 + i)^{5} (1 + i)^{3})

= ln ((1 + i)^{3} (1 + i)^{5})

(1 + i)^{3} (1 + i)^{5} = 16

= ln (16)

Rewrite

16

in power-base form:

16 = 2^{4}

= ln (2^{4})

Apply log rule:

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

ln (2^{4}) = 4 ln (2)

= 4 ln (2)

ln (6 e^{r t})

ln (a) - 2 ln (b) + 3 ln (c)

ln (x^{3}) + 2 ln (x^{2})

30 lo g_{16} (- 20 lo g_{16} (x))

lo g_{e} (2 e)