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

 
 
 
 Apr 5, 2015

Best Answer 

 #2
avatar+130477 
+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 =  

 [about  0.954434]

 

  

 Apr 5, 2015
 #1
avatar+118703 
+10

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)) =

 

(38Log2(38))+(58Log2(58))=18[3Log2(38)+5Log2(58)]=18[Log2(38)3+Log2(58)5]=18[Log2((38)3(58)5)]=18[Log2((3355)88)]=18[Log2(3355)Log288]=18[Log2(273125)8Log28]=18[Log2(84375)8Log223]=18[Log2(84375)24Log22]=18[Log2(84375)24]=3[Log2(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

 

3log2(84375)8=0.9544340029249647

 

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

 

3(log10(84375)log10(2))8=0.9544340029249649

 Apr 5, 2015
 #2
avatar+130477 
+10
Best Answer

(–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 =  

 [about  0.954434]

 

  

CPhill Apr 5, 2015

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