−(3⋅0.1log

2

​

(0.1)+3⋅0.15log

2

​

(0.15)+2⋅0.125log

2

​

(0.125))

asked by guest
on Nov 24, 2024 at 7:54 am



You asked:

Evaluate the expression: −(3⋅0.1log⁡(2)⋅0.1+3⋅0.15log⁡(2)⋅0.15+2⋅0.125log⁡(2)⋅0.125)-\left( 3 \cdot 0.1 \log\left( 2 \right) \cdot 0.1 + 3 \cdot 0.15 \log\left( 2 \right) \cdot 0.15 + 2 \cdot 0.125 \log\left( 2 \right) \cdot 0.125 \right)

MathBot Answer:

−(3⋅0.1log⁡(2)⋅0.1+3⋅0.15log⁡(2)⋅0.15+2⋅0.125log⁡(2)⋅0.125)=−103log⁡(2)800≈−0.08924269949709295858746863563774-\left( 3 \cdot 0.1 \log\left( 2 \right) \cdot 0.1 + 3 \cdot 0.15 \log\left( 2 \right) \cdot 0.15 + 2 \cdot 0.125 \log\left( 2 \right) \cdot 0.125 \right) = - \frac{103 \log{\left(2 \right)}}{800} \approx -0.08924269949709295858746863563774

−(3⋅0.1log⁡(2)⋅0.1+3⋅0.15log⁡(2)⋅0.15+2⋅0.125log⁡(2)⋅0.125)=−(3⋅110⋅log⁡(2)⋅110+3⋅15100⋅log⁡(2)⋅15100+2⋅1251000⋅log⁡(2)⋅1251000)=−(310⋅log⁡(2)⋅110+3⋅15100⋅log⁡(2)⋅15100+2⋅1251000⋅log⁡(2)⋅1251000)=−(3100⋅log⁡(2)+3⋅15100⋅log⁡(2)⋅15100+2⋅1251000⋅log⁡(2)⋅1251000)=−(3100⋅log⁡(2)+45100⋅log⁡(2)⋅15100+2⋅1251000⋅log⁡(2)⋅1251000)=−(3100⋅log⁡(2)+920⋅log⁡(2)⋅15100+2⋅1251000⋅log⁡(2)⋅1251000)=−(3100⋅log⁡(2)+274⋅log⁡(2)⋅1100+2⋅1251000⋅log⁡(2)⋅1251000)=−(3100⋅log⁡(2)+27400⋅log⁡(2)+2⋅1251000⋅log⁡(2)⋅1251000)=−(3100⋅log⁡(2)+27400⋅log⁡(2)+2501000⋅log⁡(2)⋅1251000)=−(3100⋅log⁡(2)+27400⋅log⁡(2)+14⋅log⁡(2)⋅1251000)=−(3100⋅log⁡(2)+27400⋅log⁡(2)+1254⋅log⁡(2)⋅11000)=−(3100⋅log⁡(2)+27400⋅log⁡(2)+132⋅log⁡(2))=−(39400⋅log⁡(2)+132⋅log⁡(2))=−103800⋅log⁡(2)\begin{aligned}-\left( 3 \cdot 0.1 \log\left( 2 \right) \cdot 0.1 + 3 \cdot 0.15 \log\left( 2 \right) \cdot 0.15 + 2 \cdot 0.125 \log\left( 2 \right) \cdot 0.125 \right)&=- \left(3 \cdot \frac{1}{10} \cdot \log{\left(2 \right)} \cdot \frac{1}{10} + 3 \cdot \frac{15}{100} \cdot \log{\left(2 \right)} \cdot \frac{15}{100} + 2 \cdot \frac{125}{1000} \cdot \log{\left(2 \right)} \cdot \frac{125}{1000}\right)\\&=- \left(\frac{3}{10} \cdot \log{\left(2 \right)} \cdot \frac{1}{10} + 3 \cdot \frac{15}{100} \cdot \log{\left(2 \right)} \cdot \frac{15}{100} + 2 \cdot \frac{125}{1000} \cdot \log{\left(2 \right)} \cdot \frac{125}{1000}\right)\\&=- \left(\frac{3}{100} \cdot \log{\left(2 \right)} + 3 \cdot \frac{15}{100} \cdot \log{\left(2 \right)} \cdot \frac{15}{100} + 2 \cdot \frac{125}{1000} \cdot \log{\left(2 \right)} \cdot \frac{125}{1000}\right)\\&=- \left(\frac{3}{100} \cdot \log{\left(2 \right)} + \frac{45}{100} \cdot \log{\left(2 \right)} \cdot \frac{15}{100} + 2 \cdot \frac{125}{1000} \cdot \log{\left(2 \right)} \cdot \frac{125}{1000}\right)\\&=- \left(\frac{3}{100} \cdot \log{\left(2 \right)} + \frac{9}{20} \cdot \log{\left(2 \right)} \cdot \frac{15}{100} + 2 \cdot \frac{125}{1000} \cdot \log{\left(2 \right)} \cdot \frac{125}{1000}\right)\\&=- \left(\frac{3}{100} \cdot \log{\left(2 \right)} + \frac{27}{4} \cdot \log{\left(2 \right)} \cdot \frac{1}{100} + 2 \cdot \frac{125}{1000} \cdot \log{\left(2 \right)} \cdot \frac{125}{1000}\right)\\&=- \left(\frac{3}{100} \cdot \log{\left(2 \right)} + \frac{27}{400} \cdot \log{\left(2 \right)} + 2 \cdot \frac{125}{1000} \cdot \log{\left(2 \right)} \cdot \frac{125}{1000}\right)\\&=- \left(\frac{3}{100} \cdot \log{\left(2 \right)} + \frac{27}{400} \cdot \log{\left(2 \right)} + \frac{250}{1000} \cdot \log{\left(2 \right)} \cdot \frac{125}{1000}\right)\\&=- \left(\frac{3}{100} \cdot \log{\left(2 \right)} + \frac{27}{400} \cdot \log{\left(2 \right)} + \frac{1}{4} \cdot \log{\left(2 \right)} \cdot \frac{125}{1000}\right)\\&=- \left(\frac{3}{100} \cdot \log{\left(2 \right)} + \frac{27}{400} \cdot \log{\left(2 \right)} + \frac{125}{4} \cdot \log{\left(2 \right)} \cdot \frac{1}{1000}\right)\\&=- \left(\frac{3}{100} \cdot \log{\left(2 \right)} + \frac{27}{400} \cdot \log{\left(2 \right)} + \frac{1}{32} \cdot \log{\left(2 \right)}\right)\\&=- \left(\frac{39}{400} \cdot \log{\left(2 \right)} + \frac{1}{32} \cdot \log{\left(2 \right)}\right)\\&=- \frac{103}{800} \cdot \log{\left(2 \right)}\end{aligned}

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