What is ln|1+e^{8In}|-ln(3) ?

The solution to ln|1+e^{8In}|-ln(3) is ln((|1+e^{8In}|)/3)

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simplify ln|1+e^{8In}|-ln(3)

simplify

ln 1 + e^{8 I n} - ln (3)

ln (\frac{1 + e ^{8 I n}}{3})

Solution steps

ln 1 + e^{8 I n} - ln (3)

Apply log rule:

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

ln 1 + e^{8 n I} - ln (3) = ln (\frac{e ^{8 n I} + 1}{3})

= ln (\frac{1 + e ^{8 I n}}{3})

ln (\frac{0.015}{0.025})

lo g_{8} (z x \cdot y)

\frac{1}{3} (ln (11) - ln (4))

ln (x + 1) - 2 ln (2 x^{2} + 2)

5 ln (x - 4) - ln (x^{2} - 9 x + 20) + 2 ln (x - 5)

What is ln|1+e^{8In}|-ln(3) ?
The solution to ln|1+e^{8In}|-ln(3) is ln((|1+e^{8In}|)/3)