What is 1/12 ln|(12*5)+150|-1/12 ln|12+6| ?

The solution to 1/12 ln|(12*5)+150|-1/12 ln|12+6| is 1/12 ln(35/3)

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simplify 1/12 ln|(12*5)+150|-1/12 ln|12+6|

simplify

\frac{1}{12} ln ∣ (12 \cdot 5) + 150 ∣ - \frac{1}{12} ln ∣ 12 + 6 ∣

\frac{1}{12} ln (\frac{35}{3})

Decimal Notation

0.20472 \dots

Solution steps

\frac{1}{12} ln ∣ (12 \cdot 5) + 150 ∣ - \frac{1}{12} ln ∣ 12 + 6 ∣

Apply log rule:

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

\frac{1}{12} ln ∣ (12 \cdot 5) + 150 ∣ = ln (∣ (12 \cdot 5) + 150 ∣^{\frac{1}{12}})

= ln (∣ (12 \cdot 5) + 150 ∣^{\frac{1}{12}}) - \frac{1}{12} ln ∣ 12 + 6 ∣

Apply log rule:

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

\frac{1}{12} ln ∣ 12 + 6 ∣ = ln (∣ 12 + 6 ∣^{\frac{1}{12}})

= ln (∣ (12 \cdot 5) + 150 ∣^{\frac{1}{12}}) - ln (∣ 12 + 6 ∣^{\frac{1}{12}})

Apply log rule:

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

ln (∣ (12 \cdot 5) + 150 ∣^{\frac{1}{12}}) - ln (∣ 12 + 6 ∣^{\frac{1}{12}}) = ln (\frac{∣ 12 \cdot 5 + 150 ∣ ^{\frac{1}{12}}}{∣ 12 + 6 ∣ ^{\frac{1}{12}}})

= ln (\frac{∣ 12 \cdot 5 + 150 ∣ ^{\frac{1}{12}}}{∣ 12 + 6 ∣ ^{\frac{1}{12}}})

Rewrite as

= ln ((\frac{35}{3})^{\frac{1}{12}})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= \frac{1}{12} ln (\frac{35}{3})

\frac{3 ln ( 2 ) - ln ( 3 )}{4}

2 lo g_{a} (y - 4) - 5 lo g_{a} (y + 7)

, ln ((2 π (θ + 1))^{- \frac{n}{2}})

ln (e^{4 x - 3})

ln (\frac{0.25}{0.125})

What is 1/12 ln|(12*5)+150|-1/12 ln|12+6| ?
The solution to 1/12 ln|(12*5)+150|-1/12 ln|12+6| is 1/12 ln(35/3)