SOLUTION: If log-a 4=1.386, log-a 5=1.609, and log-a 6=1.792 find the following:
log-a 2/3
log-a 20
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Question 447446: If log-a 4=1.386, log-a 5=1.609, and log-a 6=1.792 find the following:
log-a 2/3
log-a 20
Found 2 solutions by ClassPad, algebrahelp101:
Answer by ClassPad(3) (Show Source): You can put this solution on YOUR website!
Given: log-a 4=1.386, we have:
log-a 2^2=1.386
2*log-a 2=1.386
log-a 2=1.386/2 -> log-a 2=0.693
Given log-a 5=1.609 -> log-a 5=1.609 (5 is prime)
Given log-a 6=1.792, we have:
log-a (2*3)=1.792
log-a 2 + log-a 3=1.792
log-a 3 = 1.792 - log-a 2
log-a 3 = 1.792 - 0.693 -> log-a 3 = 1.099
Find:
log-a 2/3 = log-a 2 - log-a 3
= 0.693 - 1.099
= -0.406
Find:
log-a 20 = log-a (4*5)
= log-a 4 + log-a 5
= log-a 2^2 + log-a 5
= 2*log-a 2 + log-a 5
= 2*0.693 + 1.609
= 2.995
Answer by algebrahelp101(11) (Show Source): You can put this solution on YOUR website!
Given: log-a 4=1.386, we have:
log-a 2^2=1.386
2*log-a 2=1.386
log-a 2=1.386/2 -> log-a 2=0.693
Given log-a 5=1.609 -> log-a 5=1.609 (5 is prime)
Given log-a 6=1.792, we have:
log-a (2*3)=1.792
log-a 2 + log-a 3=1.792
log-a 3 = 1.792 - log-a 2
log-a 3 = 1.792 - 0.693 -> log-a 3 = 1.099
Find:
log-a 2/3 = log-a 2 - log-a 3
= 0.693 - 1.099
= -0.406
Find:
log-a 20 = log-a (4*5)
= log-a 4 + log-a 5
= log-a 2^2 + log-a 5
= 2*log-a 2 + log-a 5
= 2*0.693 + 1.609
= 2.995
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