What is 1/4 ln(2+e^{3x})-5/12 ln(2)-15 x/(12)+c ?

The solution to 1/4 ln(2+e^{3x})-5/12 ln(2)-15 x/(12)+c is ln((2^{7/12}(2+e^{3x})^{1/4})/2)-(5x)/4+c

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simplify 1/4 ln(2+e^{3x})-5/12 ln(2)-15 x/(12)+c

simplify

\frac{1}{4} ln (2 + e^{3 x}) - \frac{5}{12} ln (2) - 15 \frac{x}{12} + c

ln \frac{2 ^{\frac{7}{12}} ( 2 + e ^{3 x} ) ^{\frac{1}{4}}}{2} - \frac{5 x}{4} + c

Solution steps

\frac{1}{4} ln (2 + e^{3 x}) - \frac{5}{12} ln (2) - 15 \cdot \frac{x}{12} + c

Apply log rule:

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

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

= ln ((e^{3 x} + 2)^{\frac{1}{4}}) - \frac{5}{12} ln (2) - 15 \cdot \frac{x}{12} + c

Apply log rule:

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

\frac{5}{12} ln (2) = ln (2^{\frac{5}{12}})

= ln ((e^{3 x} + 2)^{\frac{1}{4}}) - ln (2^{\frac{5}{12}}) - 15 \cdot \frac{x}{12} + c

Apply log rule:

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

ln ((2 + e^{3 x})^{\frac{1}{4}}) - ln (2^{\frac{5}{12}}) = ln \frac{( e ^{3 x} + 2 ) ^{\frac{1}{4}}}{2 ^{\frac{5}{12}}}

= ln \frac{( e ^{3 x} + 2 ) ^{\frac{1}{4}}}{2 ^{\frac{5}{12}}} - 15 \cdot \frac{x}{12} + c

Simplify

\frac{( 2 + e ^{3 x} ) ^{\frac{1}{4}}}{2 ^{\frac{5}{12}}} : \frac{2 ^{\frac{7}{12}} ( 2 + e ^{3 x} ) ^{\frac{1}{4}}}{2}

= ln \frac{2 ^{\frac{7}{12}} ( 2 + e ^{3 x} ) ^{\frac{1}{4}}}{2} - 15 \cdot \frac{x}{12} + c

15 \cdot \frac{x}{12} = \frac{5 x}{4}

= ln \frac{2 ^{\frac{7}{12}} ( e ^{3 x} + 2 ) ^{\frac{1}{4}}}{2} - \frac{5 x}{4} + c

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What is 1/4 ln(2+e^{3x})-5/12 ln(2)-15 x/(12)+c ?
The solution to 1/4 ln(2+e^{3x})-5/12 ln(2)-15 x/(12)+c is ln((2^{7/12}(2+e^{3x})^{1/4})/2)-(5x)/4+c