What is 5/4 ln(x)-7/(2x)-5/4 ln(x+2)+C ?

The solution to 5/4 ln(x)-7/(2x)-5/4 ln(x+2)+C is ln(\sqrt[4]{(x^5)/((x+2)^5)})-7/(2x)+C

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simplify 5/4 ln(x)-7/(2x)-5/4 ln(x+2)+C

simplify

\frac{5}{4} ln (x) - \frac{7}{2 x} - \frac{5}{4} ln (x + 2) + C

Solution steps

\frac{5}{4} ln (x) - \frac{7}{2 x} - \frac{5}{4} ln (x + 2) + C

Apply log rule:

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

\frac{5}{4} ln (x) = ln (x^{\frac{5}{4}})

= ln (x^{\frac{5}{4}}) - \frac{7}{2 x} - \frac{5}{4} ln (x + 2) + C

Apply log rule:

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

\frac{5}{4} ln (x + 2) = ln ((x + 2)^{\frac{5}{4}})

= ln (x^{\frac{5}{4}}) - \frac{7}{2 x} - ln ((x + 2)^{\frac{5}{4}}) + C

Apply log rule:

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

ln (x^{\frac{5}{4}}) - ln ((x + 2)^{\frac{5}{4}}) = ln (\frac{x ^{\frac{5}{4}}}{( x + 2 ) ^{\frac{5}{4}}})

= ln (\frac{x ^{\frac{5}{4}}}{( x + 2 ) ^{\frac{5}{4}}}) - \frac{7}{2 x} + C

Combine same powers :

lo g_{10} ((tan (x))^{\circ}) 2 x

2 lo g_{10} (x) - 5 lo g_{10} (x - 1)

- lo g_{10} (2.7 \cdot 1 0^{- 7})

ln (\frac{3}{2}) + 1

lo g_{4} (\frac{4 a}{b ^{3}})

What is 5/4 ln(x)-7/(2x)-5/4 ln(x+2)+C ?
The solution to 5/4 ln(x)-7/(2x)-5/4 ln(x+2)+C is ln(\sqrt[4]{(x^5)/((x+2)^5)})-7/(2x)+C