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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (-ln|1-3|+2ln|1-2|)+(-ln|0-3|+2ln|0-2|)

simplify

(- ln ∣ 1 - 3 ∣ + 2 ln ∣ 1 - 2 ∣) + (- ln ∣ 0 - 3 ∣ + 2 ln ∣ 0 - 2 ∣)

ln (\frac{2}{3})

Decimal Notation

- 0.40546 \dots

Solution steps

(- ln ∣ 1 - 3 ∣ + 2 ln ∣ 1 - 2 ∣) + (- ln ∣ 0 - 3 ∣ + 2 ln ∣ 0 - 2 ∣)

Apply log rule:

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

2 ln ∣ 0 - 2 ∣ = ln (∣ 0 - 2 ∣^{2})

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

Apply log rule:

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

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

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

\frac{∣ 0 - 2 ∣ ^{2}}{∣ 0 - 3 ∣} = \frac{4}{3}

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

Apply log rule:

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

ln (\frac{4}{3}) - ln ∣ 1 - 3 ∣ = ln (\frac{\frac{4}{3}}{∣ 1 - 3 ∣})

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

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

= ln (\frac{2}{3}) + 2 ln ∣ 1 - 2 ∣

2 ln ∣ 1 - 2 ∣ = 0

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

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

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

lo g_{2} (76.8) - lo g_{2} (1.2)

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

ln ((\frac{10 x}{3 y})^{2})

lo g_{4} (\frac{x}{5})

- lo g_{10} (6.31 \cdot 1 0^{- 6})

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