What is 3ln(x+1)+2ln(6-x)+1/3 ln(2x+1) ?

The solution to 3ln(x+1)+2ln(6-x)+1/3 ln(2x+1) is ln((x+1)^3(6-x)^2(2x+1)^{1/3})

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simplify 3ln(x+1)+2ln(6-x)+1/3 ln(2x+1)

simplify

3 ln (x + 1) + 2 ln (6 - x) + \frac{1}{3} ln (2 x + 1)

ln ((x + 1)^{3} (6 - x)^{2} (2 x + 1)^{\frac{1}{3}})

Solution steps

3 ln (x + 1) + 2 ln (6 - x) + \frac{1}{3} ln (2 x + 1)

Apply log rule:

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

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

= ln ((x + 1)^{3}) + 2 ln (- x + 6) + \frac{1}{3} ln (2 x + 1)

Apply log rule:

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

2 ln (- x + 6) = ln ((- x + 6)^{2})

= ln ((x + 1)^{3}) + ln ((- x + 6)^{2}) + \frac{1}{3} ln (2 x + 1)

Apply log rule:

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

ln ((x + 1)^{3}) + ln ((6 - x)^{2}) = ln ((x + 1)^{3} (- x + 6)^{2})

= ln ((x + 1)^{3} (6 - x)^{2}) + \frac{1}{3} ln (2 x + 1)

Apply log rule:

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

\frac{1}{3} ln (2 x + 1) = ln ((2 x + 1)^{\frac{1}{3}})

= ln ((x + 1)^{3} (- x + 6)^{2}) + ln ((2 x + 1)^{\frac{1}{3}})

Apply log rule:

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

ln ((x + 1)^{3} (6 - x)^{2}) + ln ((2 x + 1)^{\frac{1}{3}}) = ln ((x + 1)^{3} (- x + 6)^{2} (2 x + 1)^{\frac{1}{3}})

= ln ((x + 1)^{3} (6 - x)^{2} (2 x + 1)^{\frac{1}{3}})

5 ln (2 e)

ln (22) + ln (2)

9 lo g_{5} (x) + lo g_{5} (y)

5 ln (3 x)

\frac{1}{2} lo g_{a} (- 3 lo g_{b} (x))

What is 3ln(x+1)+2ln(6-x)+1/3 ln(2x+1) ?
The solution to 3ln(x+1)+2ln(6-x)+1/3 ln(2x+1) is ln((x+1)^3(6-x)^2(2x+1)^{1/3})