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How to solve loga14 given loga2=3.01, loga6=.778 and loga7 is .845

Guest Aug 31, 2017
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loga14

Given  loga2=3.01, loga6=.778 and loga7 is .845

 

Note that we only need   loga 2 = 3.01   and loga 7  = .845     to solve this

 

By a log property......log m  + log n  =  log ( m * n)    ....so.....

 

loga 2  +   loga 7  =     loga  ( 2 * 7 )  =   loga (14)

 

3.01    +   .845   =    loga (14)   =   3.855

 

BTW -  "a"  couldn't be a "real" base!!!

 

 

cool cool cool

CPhill  Aug 31, 2017
edited by CPhill  Aug 31, 2017

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