What is 1/3 [2ln(x+3)+ln(x)-in(x^2-1)] ?

The solution to 1/3 [2ln(x+3)+ln(x)-in(x^2-1)] is 1/3 [ln((x+3)^2x)-in(x^2-1)]

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simplify 1/3 [2ln(x+3)+ln(x)-in(x^2-1)]

simplify

\frac{1}{3} [2 ln (x + 3) + ln (x) - in (x^{2} - 1)]

\frac{1}{3} [ln ((x + 3)^{2} x) - in (x^{2} - 1)]

Solution steps

\frac{1}{3} [2 ln (x + 3) + ln (x) - in (x^{2} - 1)]

Apply log rule:

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

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

= \frac{1}{3} [ln ((x + 3)^{2}) + ln (x) - in (x^{2} - 1)]

Apply log rule:

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

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

= \frac{1}{3} [ln ((x + 3)^{2} x) - in (x^{2} - 1)]

\frac{2}{3} lo g_{10} (x) - \frac{1}{3} lo g_{10} (y)

ln (\frac{1 - ( - 1 )}{1 + 2})

- lo g_{10} (5.08 \cdot 1 0^{- 5})

2.45 + lo g_{10} (\frac{0.047}{0.054})

ln (x) + \frac{2}{5} ln (5 u^{2} + 1)

What is 1/3 [2ln(x+3)+ln(x)-in(x^2-1)] ?
The solution to 1/3 [2ln(x+3)+ln(x)-in(x^2-1)] is 1/3 [ln((x+3)^2x)-in(x^2-1)]