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I'd be grateful if anyone could explain me the way of solving, for me it turns out to be 3 instead to 2

log(3,6)*log(9,6)+log(4,6)*log(18,6)

Guest Aug 1, 2014

Best Answer 

 #3
avatar+90988 
+16

$$log_63*log_69+log_64*log_618\\\\
=log_63*2log_63+log_64*log_6(9*2)\\\\
=log_63*2log_63+log_64*(log_69+log_62)\\\\
=log_63*2log_63+2log_62*(2log_63+log_62)\\\\
=log_63*2log_63+4log_62log_63+2(log_62)^2\\\\
=2(log_63)^2+4log_62log_63+2(log_62)^2\\\\
=2[(log_63)^2+2log_62log_63+(log_62)^2]\\\\
=2[(log_63+log_62)^2]\\\\
=2[(log_66)^2]\\\\
=2[1^2]\\\\
=2\\$$

Melody  Aug 1, 2014
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5+0 Answers

 #1
avatar+4471 
+13

log(3,6)*log(9,6)+log(4,6)*log(18,6) =

log(3,6)*[log(3,6) + log(3,6)] + log(4,6)*[log(3,6) + log(3,6) + log(2,6)]

log(3,6) = ~0.613, log(4,6) = ~0.774, & log(2,6) = ~0.387.

So, 0.613*[0.613 + 0.613] + 0.774*[0.613 + 0.613 + 0.387]

= 0.752 + 1.248 

2.

AzizHusain  Aug 1, 2014
 #2
avatar+78574 
+3

Sorry, for the re-post of Aziz's excellent answer....I was confused by the notation in this question and I wanted to get some "practice" in to make sure I understood it !!!......I'll give Aziz full credit, here!!!

-------------------------------------------------------------------------------------------------------------------------

 

log(3)/log(6) = 0.6131471927654584

log(9)/log(6) = 1.2262943855309169

So

[log(3)/log(6)]*[log(9)/log(6)] = 0.75189895999232446469782360080696

And

log(4)/log(6) = 0.7737056144690832

log(18)/log(6) = 1.6131471927654584

So

[log(4)/log(6)]*[log(18)/log(6)] = 1.24810104000767559661689567573888

So we have

0.75189895999232446469782360080696 +1.24810104000767559661689567573888 =

2.00000000000000006131471927654584 ≈  2

CPhill  Aug 1, 2014
 #3
avatar+90988 
+16
Best Answer

$$log_63*log_69+log_64*log_618\\\\
=log_63*2log_63+log_64*log_6(9*2)\\\\
=log_63*2log_63+log_64*(log_69+log_62)\\\\
=log_63*2log_63+2log_62*(2log_63+log_62)\\\\
=log_63*2log_63+4log_62log_63+2(log_62)^2\\\\
=2(log_63)^2+4log_62log_63+2(log_62)^2\\\\
=2[(log_63)^2+2log_62log_63+(log_62)^2]\\\\
=2[(log_63+log_62)^2]\\\\
=2[(log_66)^2]\\\\
=2[1^2]\\\\
=2\\$$

Melody  Aug 1, 2014
 #4
avatar+78574 
0

Very nice, Melody....that is well laid out !!!!

3 points from me!!!

 

CPhill  Aug 1, 2014
 #5
avatar+90988 
0

Thanks Chris. 

Melody  Aug 1, 2014

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