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What is 1/3(log4x + log4z) as a single logarithm

 Nov 14, 2014

Best Answer 

 #2
avatar+23254 
+5

When you wrote log4x, did you mean:  log (to the base 4) of x  or log (to the base to) of 4x?

First:  log (to the base 4) of x:  I'll skip the '4' when I type:

   (1/3)(logx + logz)  =  (1/3)(log(xy)      (because adding logs means that the numbers were multiplied)

    =  log[ (xy)^(1/3) ]       (because a multiplied goes into a log as an exponent) 

Second:  log (to the base 10 )

   (1/3)[log(4x) + log(4y)]  =  (1/3)[log(4x·4y)]  =  (1/3)[log(16xy)]  =  log[ (16xy)^(1/3) ]

 Nov 14, 2014
 #1
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this person asked this question so long ago while others who asked after them got a response yet this person didn't what terrible service

 Nov 14, 2014
 #2
avatar+23254 
+5
Best Answer

When you wrote log4x, did you mean:  log (to the base 4) of x  or log (to the base to) of 4x?

First:  log (to the base 4) of x:  I'll skip the '4' when I type:

   (1/3)(logx + logz)  =  (1/3)(log(xy)      (because adding logs means that the numbers were multiplied)

    =  log[ (xy)^(1/3) ]       (because a multiplied goes into a log as an exponent) 

Second:  log (to the base 10 )

   (1/3)[log(4x) + log(4y)]  =  (1/3)[log(4x·4y)]  =  (1/3)[log(16xy)]  =  log[ (16xy)^(1/3) ]

geno3141 Nov 14, 2014

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