Solve =-[2/10b_{2}(2/10)+8/10log_{2}(8/10)]

$$= - [ \frac { 2 } { 10 } b _ { 2 } ( \frac { 2 } { 10 } ) + \frac { 8 } { 10 } \log _ { 2 } ( \frac { 8 } { 10 } ) ]$$

Evaluate

$$-\frac{b_{2}}{25}+\frac{4\log_{2}\left(5\right)}{5}-\frac{8}{5}$$

Solution Steps

Multiply \frac{2}{10} and \frac{2}{10} to get \left(\frac{2}{10}\right)^{2}.

$$-\left(\left(\frac{2}{10}\right)^{2}b_{2}+\frac{8}{10}\log_{2}\left(\frac{8}{10}\right)\right)$$

Reduce the fraction \frac{2}{10} to lowest terms by extracting and canceling out 2.

$$-\left(\left(\frac{1}{5}\right)^{2}b_{2}+\frac{8}{10}\log_{2}\left(\frac{8}{10}\right)\right)$$

Calculate \frac{1}{5} to the power of 2 and get \frac{1}{25}.

$$-\left(\frac{1}{25}b_{2}+\frac{8}{10}\log_{2}\left(\frac{8}{10}\right)\right)$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$-\left(\frac{1}{25}b_{2}+\frac{4}{5}\log_{2}\left(\frac{8}{10}\right)\right)$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$-\left(\frac{1}{25}b_{2}+\frac{4}{5}\log_{2}\left(\frac{4}{5}\right)\right)$$

To find the opposite of \frac{1}{25}b_{2}+\frac{4}{5}\log_{2}\left(\frac{4}{5}\right), find the opposite of each term.

$$-\frac{1}{25}b_{2}-\frac{4}{5}\log_{2}\left(\frac{4}{5}\right)$$

Factor

$$\frac{-b_{2}+20\log_{2}\left(5\right)-40}{25}$$

Solution Steps

Consider \frac{1}{5}\times \frac{1}{5}b_{2}+\frac{4}{5}\ln(\frac{4}{5})\ln(2)^{-1}. Factor out \frac{1}{25}.

$$\frac{b_{2}+20\log_{2}\left(\frac{4}{5}\right)}{25}$$

Rewrite the complete factored expression. Simplify.

$$-\frac{b_{2}+20\log_{2}\left(\frac{4}{5}\right)}{25}$$