What is 2ln(y)+9/5 ln(y+2)+1/5 ln(y-3) ?

The solution to 2ln(y)+9/5 ln(y+2)+1/5 ln(y-3) is ln(y^2(y+2)^{9/5}(y-3)^{1/5})

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simplify 2ln(y)+9/5 ln(y+2)+1/5 ln(y-3)

simplify

2 ln (y) + \frac{9}{5} ln (y + 2) + \frac{1}{5} ln (y - 3)

ln (y^{2} (y + 2)^{\frac{9}{5}} (y - 3)^{\frac{1}{5}})

Solution steps

2 ln (y) + \frac{9}{5} ln (y + 2) + \frac{1}{5} ln (y - 3)

Apply log rule:

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

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

= ln (y^{2}) + \frac{9}{5} ln (y + 2) + \frac{1}{5} ln (y - 3)

Apply log rule:

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

\frac{9}{5} ln (y + 2) = ln ((y + 2)^{\frac{9}{5}})

= ln (y^{2}) + ln ((y + 2)^{\frac{9}{5}}) + \frac{1}{5} ln (y - 3)

Apply log rule:

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

ln (y^{2}) + ln ((y + 2)^{\frac{9}{5}}) = ln (y^{2} (y + 2)^{\frac{9}{5}})

= ln (y^{2} (y + 2)^{\frac{9}{5}}) + \frac{1}{5} ln (y - 3)

Apply log rule:

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

\frac{1}{5} ln (y - 3) = ln ((y - 3)^{\frac{1}{5}})

= ln (y^{2} (y + 2)^{\frac{9}{5}}) + ln ((y - 3)^{\frac{1}{5}})

Apply log rule:

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

ln (y^{2} (y + 2)^{\frac{9}{5}}) + ln ((y - 3)^{\frac{1}{5}}) = ln (y^{2} (y + 2)^{\frac{9}{5}} (y - 3)^{\frac{1}{5}})

= ln (y^{2} (y + 2)^{\frac{9}{5}} (y - 3)^{\frac{1}{5}})

ln (\frac{80}{9996.6})

lo g_{10} (\frac{4 x ^{6}}{- 1})

2 lo g_{8} (z) - \frac{1}{2} lo g_{8} (w) + 4 lo g_{8} (x)

2^{\frac{π}{2}} lo g_{10} (2)

- \frac{1}{8} (ln a + \frac{5}{2} - ln a + \frac{1}{2}) + C

What is 2ln(y)+9/5 ln(y+2)+1/5 ln(y-3) ?
The solution to 2ln(y)+9/5 ln(y+2)+1/5 ln(y-3) is ln(y^2(y+2)^{9/5}(y-3)^{1/5})