(–3/8 Log2(3/8)) + (– 5/8 Log2(5/8))

(–3/8 Log2(3/8)) + (– 5/8 Log2(5/8))=

(-3/8) [log₂ 3 - log₂ 8] + (-5/8) [log₂ 5 - log₂  8] =

(-3/8) [log₂ 3  - 3] + (-5/8) [ log₂ 5  - 3] =

-3 [ -3/8 -5/8]  - [ (3/8) log₂ 3 + (5/8)  log₂ 5 ] =

3  - [ 1/8)] * [ 3 log₂ 3 + 5 log₂ 5 ]  =

3 - [1/8] * [ log₂ 3^3 +  log₂ 5^5] =

3 - [1/8] * [ log₂ 27 + log₂ 3125] =

3 - [1/8] * [log₂ (27 * 3125)] =

3 - [log₂ 84375] / 8 =

[24 -  log₂ 84375 ] / 8 =  

 [about  0.954434]

This answer has been modified, 

Thanks CPhill and Alan for letting me know of my careless error :))

(–3/8 Log2(3/8)) + (– 5/8 Log2(5/8)) =

$$\\(-\frac{3}{8}Log_2(\frac{3}{8})) + (-\frac{5}{8} Log_2(\frac{5}{8})) \\\\
=-\frac{1}{8}[3Log_2(\frac{3}{8}) + 5 Log_2(\frac{5}{8})] \\\\
=-\frac{1}{8}[Log_2(\frac{3}{8})^3 + Log_2(\frac{5}{8})^5] \\\\
=-\frac{1}{8}[Log_2((\frac{3}{8})^3*(\frac{5}{8})^5 )] \\\\
=-\frac{1}{8}[Log_2(\frac{(3^3*5^5)}{8^8})] \\\\
=-\frac{1}{8}[Log_2 (3^3*5^5)-Log_2 8^8] \\\\
=-\frac{1}{8}[Log_2 (27*3125)-8Log_2 8] \\\\
=-\frac{1}{8}[Log_2 (84375)-8Log_2 2^3]\\\\
=-\frac{1}{8}[Log_2 (84375)-24Log_2 2]\\\\
=-\frac{1}{8}[Log_2 (84375)-24]\\\\
=3-\frac{[Log_2 (84375)]}{8}\\\\$$

This is the same as Chris's answer below.  :)

The question is finished but if you want an approximation ...

 this can be entered into the web 2 calc like this

3-log(84375,2)/8

$${\mathtt{3}}{\mathtt{\,-\,}}{\frac{{{log}}_{{\mathtt{2}}}{\left({\mathtt{84\,375}}\right)}}{{\mathtt{8}}}} = {\mathtt{0.954\: \!434\: \!002\: \!924\: \!964\: \!7}}$$

Most calcs only do base 10 or base e so this is how you would do it on one of those (uaing change of base rule)

3-(log(84375)/log(2))/8

$${\mathtt{3}}{\mathtt{\,-\,}}{\frac{\left({\frac{{log}_{10}\left({\mathtt{84\,375}}\right)}{{log}_{10}\left({\mathtt{2}}\right)}}\right)}{{\mathtt{8}}}} = {\mathtt{0.954\: \!434\: \!002\: \!924\: \!964\: \!9}}$$