What is 345(252*ln(252)+(1-252)*ln(1-252)) ?

The solution to 345(252*ln(252)+(1-252)*ln(1-252)) is 86940ln(-(252^{252})/(251^{251)})

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simplify 345(252ln(252)+(1-252)ln(1-252))

simplify

345 (252 \cdot ln (252) + (1 - 252) \cdot ln (1 - 252))

86940 ln (- \frac{25 2 ^{252}}{25 1 ^{251}})

Solution steps

345 (252 ln (252) + (1 - 252) ln (1 - 252))

Apply log rule:

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

ln (1 - 252) (1 - 252) = ln ((1 - 252)^{(1 - 252)})

= 345 (252 ln (252) + ln ((1 - 252)^{(1 - 252)}))

Apply log rule:

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

ln (252) + ln ((1 - 252)^{(1 - 252)}) = ln (252 (1 - 252)^{1 - 252})

= 345 \cdot 252 ln (252 (1 - 252)^{1 - 252})

Simplify

25 2^{252} (1 - 252)^{(1 - 252)} : - \frac{25 2 ^{252}}{25 1 ^{251}}

= 345 \cdot 252 ln (- \frac{25 2 ^{252}}{25 1 ^{251}})

Multiply the numbers:

345 \cdot 252 = 86940

= 86940 ln (- \frac{25 2 ^{252}}{25 1 ^{251}})

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4 lo g_{10} (a) - 5 lo g_{10} (b)

What is 345(252*ln(252)+(1-252)*ln(1-252)) ?
The solution to 345(252*ln(252)+(1-252)*ln(1-252)) is 86940ln(-(252^{252})/(251^{251)})