What is 1/3 ln|3y+1|+C-1/4 ln|4x-3|+C ?

The solution to 1/3 ln|3y+1|+C-1/4 ln|4x-3|+C is ln((|3y+1|^{1/3}|4x-3|^{3/4})/(|4x-3|))+2C

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simplify 1/3 ln|3y+1|+C-1/4 ln|4x-3|+C

simplify

\frac{1}{3} ln ∣ 3 y + 1 ∣ + C - \frac{1}{4} ln ∣ 4 x - 3 ∣ + C

ln (\frac{∣ 3 y + 1 ∣ ^{\frac{1}{3}} ∣ 4 x - 3 ∣ ^{\frac{3}{4}}}{∣ 4 x - 3 ∣}) + 2 C

Solution steps

\frac{1}{3} ln ∣ 3 y + 1 ∣ + C - \frac{1}{4} ln ∣ 4 x - 3 ∣ + C

Apply log rule:

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

\frac{1}{3} ln ∣ 3 y + 1 ∣ = ln (∣ 3 y + 1 ∣^{\frac{1}{3}})

= ln (∣ 3 y + 1 ∣^{\frac{1}{3}}) + C - \frac{1}{4} ln ∣ 4 x - 3 ∣ + C

Apply log rule:

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

\frac{1}{4} ln ∣ 4 x - 3 ∣ = ln (∣ 4 x - 3 ∣^{\frac{1}{4}})

= ln (∣ 3 y + 1 ∣^{\frac{1}{3}}) + C - ln (∣ 4 x - 3 ∣^{\frac{1}{4}}) + C

Apply log rule:

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

ln (∣ 3 y + 1 ∣^{\frac{1}{3}}) - ln (∣ 4 x - 3 ∣^{\frac{1}{4}}) = ln (\frac{∣ 3 y + 1 ∣ ^{\frac{1}{3}}}{∣ 4 x - 3 ∣ ^{\frac{1}{4}}})

= ln (\frac{∣ 3 y + 1 ∣ ^{\frac{1}{3}}}{∣ 4 x - 3 ∣ ^{\frac{1}{4}}}) + C + C

Simplify

\frac{∣ 3 y + 1 ∣ ^{\frac{1}{3}}}{∣ 4 x - 3 ∣ ^{\frac{1}{4}}} : \frac{∣ 3 y + 1 ∣ ^{\frac{1}{3}} ∣ 4 x - 3 ∣ ^{\frac{3}{4}}}{∣ 4 x - 3 ∣}

= ln (\frac{∣ 3 y + 1 ∣ ^{\frac{1}{3}} ∣ 4 x - 3 ∣ ^{\frac{3}{4}}}{∣ 4 x - 3 ∣}) + C + C

Add similar elements:

C + C = 2 C

= ln (\frac{∣ 3 y + 1 ∣ ^{\frac{1}{3}} ∣ 4 x - 3 ∣ ^{\frac{3}{4}}}{∣ 4 x - 3 ∣}) + 2 C

\frac{1}{2} lo g_{c} (x) + 5 lo g_{c} (y) - 3 lo g_{c} (x)

ln 2 + 3 - ln 2 + 1

lo g_{10} (\frac{x ^{2}}{z})

ln (\frac{x ^{9} x ^{2} + 3}{( x + 3 ) ^{6}})

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

What is 1/3 ln|3y+1|+C-1/4 ln|4x-3|+C ?
The solution to 1/3 ln|3y+1|+C-1/4 ln|4x-3|+C is ln((|3y+1|^{1/3}|4x-3|^{3/4})/(|4x-3|))+2C