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(–3/8 Log2(3/8)) + (– 5/8 Log2(5/8))

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

Apr 5, 2015

#2
+10

(–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) [ log5  - 3] =

-3 [ -3/8 -5/8]  - [ (3/8) log2 3 + (5/8)  log5 ] =

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 =   Apr 5, 2015

#1
+10

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

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

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Apr 5, 2015
#2
+10

(–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) [ log5  - 3] =

-3 [ -3/8 -5/8]  - [ (3/8) log2 3 + (5/8)  log5 ] =

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 =   