What is ln|1|-3/2 ln|5|+2ln|6|+C ?

The solution to ln|1|-3/2 ln|5|+2ln|6|+C is C+ln((36sqrt(5))/(25))

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simplify ln|1|-3/2 ln|5|+2ln|6|+C

simplify

ln ∣ 1 ∣ - \frac{3}{2} ln ∣ 5 ∣ + 2 ln ∣ 6 ∣ + C

C + ln (\frac{36 5}{25})

Solution steps

ln ∣ 1 ∣ - \frac{3}{2} ln ∣ 5 ∣ + 2 ln ∣ 6 ∣ + C

Apply log rule:

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

\frac{3}{2} ln ∣ 5 ∣ = ln (∣ 5 ∣^{\frac{3}{2}})

= ln ∣ 1 ∣ - ln (∣ 5 ∣^{\frac{3}{2}}) + 2 ln ∣ 6 ∣ + C

Apply log rule:

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

2 ln ∣ 6 ∣ = ln (∣ 6 ∣^{2})

= ln ∣ 1 ∣ - ln (∣ 5 ∣^{\frac{3}{2}}) + ln (∣ 6 ∣^{2}) + C

Apply log rule:

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

ln (∣ 6 ∣^{2}) - ln (∣ 5 ∣^{\frac{3}{2}}) = ln (\frac{∣ 6 ∣ ^{2}}{∣ 5 ∣ ^{\frac{3}{2}}})

= ln ∣ 1 ∣ + ln (\frac{∣ 6 ∣ ^{2}}{∣ 5 ∣ ^{\frac{3}{2}}}) + C

\frac{∣ 6 ∣ ^{2}}{∣ 5 ∣ ^{\frac{3}{2}}} = \frac{36 5}{25}

= ln ∣ 1 ∣ + ln (\frac{36 5}{25}) + C

ln ∣ 1 ∣ = 0

= 0 + ln (\frac{36 5}{25}) + C

0 + ln (\frac{36 5}{25}) + C = ln (\frac{36 5}{25}) + C

= C + ln (\frac{36 5}{25})

ln (x^{9} 5 - x)

lo g_{10} (\frac{8 m}{n})

\frac{10}{ln ( 302 )} - \frac{3000}{151 ln ^{2} ( 302 )} + 3

lo g_{2} (\frac{x}{( y + 1 ) z})

ln (x) + 8 ln (y)

What is ln|1|-3/2 ln|5|+2ln|6|+C ?
The solution to ln|1|-3/2 ln|5|+2ln|6|+C is C+ln((36sqrt(5))/(25))