=[1/(1-2.257)][ln (0.236/0.0783)+ 2.257ln((1-0.236)/(1-0.0783))]
asked by guest
on Mar 31, 2025 at 1:51 pm
You asked:
Evaluate the expression:
1−2.2571(ln(0.07830.236)+2.257ln(1−0.07831−0.236))
MathBot Answer:
1−2.2571(ln(0.07830.236)+2.257ln(1−0.07831−0.236))=−12571000log(7832360)−12572257log(92177640)≈−0.54077456952390638963504929752404
1−2.2571(ln(0.07830.236)+2.257ln(1−0.07831−0.236))=1−(2+1000257)1⋅(log(1000236⋅100007831)+(2+1000257)⋅log((1−1000236)⋅1−100007831))=1−100022571⋅(log(1000236⋅100007831)+(2+1000257)⋅log((1−1000236)⋅1−100007831))=−100012571⋅(log(1000236⋅100007831)+(2+1000257)⋅log((1−1000236)⋅1−100007831))=−12571000⋅(log(1000236⋅100007831)+(2+1000257)⋅log((1−1000236)⋅1−100007831))=−12571000⋅(log(236⋅78310)+(2+1000257)⋅log((1−1000236)⋅1−100007831))=−12571000⋅(log(7832360)+(2+1000257)⋅log((1−1000236)⋅1−100007831))=−12571000⋅(log(7832360)+10002257⋅log((1−1000236)⋅1−100007831))=−12571000⋅(log(7832360)+10002257⋅log((1−25059)⋅1−100007831))=−12571000⋅(log(7832360)+10002257⋅log(250191⋅1−100007831))=−12571000⋅(log(7832360)+10002257⋅log(250191⋅1000092171))=−12571000⋅(log(7832360)+10002257⋅log(92177640))=−12571000⋅(10002257⋅log(92177640)+log(7832360))=−12572257⋅log(92177640)−12571000⋅log(7832360)