What is 2ln(3)-kn^2-2ln(2)+ln(1) ?

The solution to 2ln(3)-kn^2-2ln(2)+ln(1) is ln(9/4)-kn^2

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simplify 2ln(3)-kn^2-2ln(2)+ln(1)

simplify

2 ln (3) - k n^{2} - 2 ln (2) + ln (1)

ln (\frac{9}{4}) - k n^{2}

Solution steps

2 ln (3) - k n^{2} - 2 ln (2) + ln (1)

Apply log rule:

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

2 ln (3) = ln (3^{2})

= ln (3^{2}) - k n^{2} - 2 ln (2) + ln (1)

Apply log rule:

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

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

= ln (3^{2}) - k n^{2} - ln (2^{2}) + ln (1)

Apply log rule:

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

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

= ln (\frac{3 ^{2}}{2 ^{2}}) - k n^{2} + ln (1)

\frac{3 ^{2}}{2 ^{2}} = \frac{9}{4}

= ln (\frac{9}{4}) - k n^{2} + ln (1)

Apply log rule:

lo g_{a} (1) = 0

= ln (\frac{9}{4}) - k n^{2} + 0

ln (\frac{9}{4}) - k n^{2} + 0 = ln (\frac{9}{4}) - k n^{2}

= ln (\frac{9}{4}) - k n^{2}

lo g_{2} (\frac{7}{11} e^{(\frac{15}{4})})

6 lo g_{5} (x) - 5 lo g_{5} (y)

ln (\frac{3 x - 5}{2 x - 1})

(\frac{1}{2} ln ∣ 2 (10) + 1 ∣ + 2 ln ∣ 10 - 1 ∣) - 5

ln (\frac{( 1 + 10 x )}{x})

What is 2ln(3)-kn^2-2ln(2)+ln(1) ?
The solution to 2ln(3)-kn^2-2ln(2)+ln(1) is ln(9/4)-kn^2