What is (ln|1-3|-ln|0-3|)-(ln|1+3|-ln|0+3|) ?

The solution to (ln|1-3|-ln|0-3|)-(ln|1+3|-ln|0+3|) is -ln(2)

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simplify (ln|1-3|-ln|0-3|)-(ln|1+3|-ln|0+3|)

simplify

(ln ∣ 1 - 3 ∣ - ln ∣ 0 - 3 ∣) - (ln ∣ 1 + 3 ∣ - ln ∣ 0 + 3 ∣)

- ln (2)

Decimal

- 0.69314 \dots

Solution steps

(ln ∣ 1 - 3 ∣ - ln ∣ 0 - 3 ∣) - (ln ∣ 1 + 3 ∣ - ln ∣ 0 + 3 ∣)

Apply log rule:

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

ln ∣ 1 - 3 ∣ - ln ∣ 0 - 3 ∣ = ln (\frac{∣ 1 - 3 ∣}{∣ 0 - 3 ∣})

= ln (\frac{∣ 1 - 3 ∣}{∣ 0 - 3 ∣}) - (ln ∣ 1 + 3 ∣ - ln ∣ 0 + 3 ∣)

\frac{∣ 1 - 3 ∣}{∣ 0 - 3 ∣} = \frac{2}{3}

= ln (\frac{2}{3}) - (ln ∣ 1 + 3 ∣ - ln ∣ 0 + 3 ∣)

Apply log rule:

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

ln ∣ 1 + 3 ∣ - ln ∣ 0 + 3 ∣ = ln (\frac{∣ 1 + 3 ∣}{∣ 0 + 3 ∣})

= ln (\frac{2}{3}) - ln (\frac{∣ 1 + 3 ∣}{∣ 0 + 3 ∣})

\frac{∣ 1 + 3 ∣}{∣ 0 + 3 ∣} = \frac{4}{3}

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

Apply log rule:

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

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

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

\frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2}

= ln (\frac{1}{2})

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln (2)

ln (\frac{c ^{- 1} \cdot e ^{(n - θ) t}}{( 1 + t )})

\frac{3}{28} ln ∣ 15 ∣ - \frac{3}{28} ln ∣ 8 ∣

- \frac{1}{12} ln ∣ y ∣ + \frac{1}{12} ln ∣ y - 3 ∣

ln (4 a^{(3)} b^{(5)})

4 lo g_{10} (x + 6) - lo g_{10} (x)

What is (ln|1-3|-ln|0-3|)-(ln|1+3|-ln|0+3|) ?
The solution to (ln|1-3|-ln|0-3|)-(ln|1+3|-ln|0+3|) is -ln(2)