What is simplify-0.1ln(0.1)-(1-0.1)ln(1-0.1) ?

The solution to simplify-0.1ln(0.1)-(1-0.1)ln(1-0.1) is -ln(0.72246…)

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simplify-0.1ln(0.1)-(1-0.1)ln(1-0.1)

simplify

- 0.1 ln (0.1) - (1 - 0.1) ln (1 - 0.1)

- ln (0.72246 \dots)

Decimal Notation

0.32508 \dots

Solution steps

- 0.1 ln (0.1) - (1 - 0.1) ln (1 - 0.1)

Apply log rule:

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

0.1 ln (0.1) = ln (0. 1^{0.1})

= - ln (0. 1^{0.1}) - ln (1 - 0.1) (1 - 0.1)

Apply log rule:

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

ln (1 - 0.1) (1 - 0.1) = ln ((1 - 0.1)^{(1 - 0.1)})

= - ln (0. 1^{0.1}) - ln ((1 - 0.1)^{(1 - 0.1)})

Rewrite

- ln (0. 1^{0.1}) - ln ((1 - 0.1)^{(1 - 0.1)})

- (ln (0. 1^{0.1}) + ln ((1 - 0.1)^{(1 - 0.1)}))

= - (ln (0. 1^{0.1}) + ln ((1 - 0.1)^{(1 - 0.1)}))

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

ln (0. 1^{0.1}) + ln ((1 - 0.1)^{(1 - 0.1)}) = ln (0. 1^{0.1} (1 - 0.1)^{1 - 0.1})

= - ln (0. 1^{0.1} (1 - 0.1)^{1 - 0.1})

0. 1^{0.1} (1 - 0.1)^{(1 - 0.1)} = 0.72246 \dots

= - ln (0.72246 \dots)

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

ln (e^{x^{2}} x)

2 ln ∣ 2 ∣ - 2 ln \frac{1}{4}

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

- lo g_{10} (3.52 x 1 0^{- 4})

What is simplify-0.1ln(0.1)-(1-0.1)ln(1-0.1) ?
The solution to simplify-0.1ln(0.1)-(1-0.1)ln(1-0.1) is -ln(0.72246…)