What is 2ln(x+1)-2ln(x)+4ln(3x+1) ?

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

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simplify 2ln(x+1)-2ln(x)+4ln(3x+1)

simplify

2 ln (x + 1) - 2 ln (x) + 4 ln (3 x + 1)

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

Solution steps

2 ln (x + 1) - 2 ln (x) + 4 ln (3 x + 1)

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

4 ln (3 x + 1) = ln ((3 x + 1)^{4})

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

Apply log rule:

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

\frac{1}{3} lo g_{10} (m) - 2 lo g_{10} (n)

ln (\frac{( x ^{2} - 5 x + 3 ) ( x - 5 ) ^{5}}{( 3 x ^{2} + 2 ) ^{6}})

20 e^{9} + ln (12) + 50 + ln (9)

4 lo g_{10} (x) + 3 lo g_{10} (y) - 2 lo g_{10} (z)

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