What is 1/4 x^2(2ln^2(3x)-2ln(3x)+1) ?

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

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simplify 1/4 x^2(2ln^2(3x)-2ln(3x)+1)

simplify

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

\frac{x ^{2}}{4}

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

= \frac{1}{4} x^{2} (ln^{2} (\frac{( 3 x ) ^{2}}{( 3 x ) ^{2}}) + 1)

Apply rule

\frac{a}{a} = 1

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= (ln^{2} (1) + 1) \frac{1 \cdot x ^{2}}{4}

Multiply:

1 \cdot x^{2} = x^{2}

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

ln^{2} (1) = 0

= (0 + 1) \frac{x ^{2}}{4}

Simplify

= \frac{x ^{2}}{4}

ln (2 + x) - ln (2)

ln (\frac{6.35 \cdot 1 0 ^{- 3}}{0.00555})

ln (\frac{e ^{4}}{19})

ln (3 x e)

- \frac{1}{2} ln ∣ (9) + 1 ∣ + \frac{1}{2} ln ∣ (9) - 1 ∣

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