What is 1.75ln|y+2|+0.25ln|y-3|+2ln|y| ?

The solution to 1.75ln|y+2|+0.25ln|y-3|+2ln|y| is ln(|y+2|^{1.75}|y-3|^{0.25}|y|^2)

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simplify 1.75ln|y+2|+0.25ln|y-3|+2ln|y|

simplify

1.75 ln ∣ y + 2 ∣ + 0.25 ln ∣ y - 3 ∣ + 2 ln ∣ y ∣

ln (∣ y + 2 ∣^{1.75} ∣ y - 3 ∣^{0.25} ∣ y ∣^{2})

Solution steps

1.75 ln ∣ y + 2 ∣ + 0.25 ln ∣ y - 3 ∣ + 2 ln ∣ y ∣

Apply log rule:

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

1.75 ln ∣ y + 2 ∣ = ln (∣ y + 2 ∣^{1.75})

= ln (∣ y + 2 ∣^{1.75}) + 0.25 ln ∣ y - 3 ∣ + 2 ln ∣ y ∣

Apply log rule:

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

0.25 ln ∣ y - 3 ∣ = ln (∣ y - 3 ∣^{0.25})

= ln (∣ y + 2 ∣^{1.75}) + ln (∣ y - 3 ∣^{0.25}) + 2 ln ∣ y ∣

Apply log rule:

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

ln (∣ y + 2 ∣^{1.75}) + ln (∣ y - 3 ∣^{0.25}) = ln (∣ y + 2 ∣^{1.75} ∣ y - 3 ∣^{0.25})

= ln (∣ y + 2 ∣^{1.75} ∣ y - 3 ∣^{0.25}) + 2 ln ∣ y ∣

Apply log rule:

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

2 ln ∣ y ∣ = ln (∣ y ∣^{2})

= ln (∣ y + 2 ∣^{1.75} ∣ y - 3 ∣^{0.25}) + ln (∣ y ∣^{2})

Apply log rule:

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

ln (∣ y + 2 ∣^{1.75} ∣ y - 3 ∣^{0.25}) + ln (∣ y ∣^{2}) = ln (∣ y ∣^{2} ∣ y + 2 ∣^{1.75} ∣ y - 3 ∣^{0.25})

= ln (∣ y + 2 ∣^{1.75} ∣ y - 3 ∣^{0.25} ∣ y ∣^{2})

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What is 1.75ln|y+2|+0.25ln|y-3|+2ln|y| ?
The solution to 1.75ln|y+2|+0.25ln|y-3|+2ln|y| is ln(|y+2|^{1.75}|y-3|^{0.25}|y|^2)