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what is the answer to:

log7 + log (n-2) = log 6n

Guest Feb 28, 2017
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3+0 Answers

 #1
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Solve for n:
log(7) + log(n - 2) = log(6 n)

Subtract log(6 n) from both sides:
log(7) + log(n - 2) - log(6 n) = 0

log(7) + log(n - 2) - log(6 n) = log(7) + log(n - 2) + log(1/(6 n)) = log((7 (n - 2))/(6 n)):
log((7 (n - 2))/(6 n)) = 0

Cancel logarithms by taking exp of both sides:
(7 (n - 2))/(6 n) = 1

Multiply both sides by 6 n:
7 (n - 2) = 6 n

Expand out terms of the left hand side:
7 n - 14 = 6 n

Subtract 6 n - 14 from both sides:
Answer: |n = 14

Guest Feb 28, 2017
 #2
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log7 + log (n-2) = log (6n)      by a log property we can write

 

log [7 ( n - 2) ]  = log (6n)       the log bases are the same, so we can solve this

 

7(n - 2)   =   6n

 

7n - 14   = 6n          add 14 to both sides, subtract 6n from both sides

 

n  = 14

 

 

cool cool cool

CPhill  Feb 28, 2017
 #3
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log7+log(n-2)=log6n             it is true that loga+logb=log(ab)

log(7(n-2))=log6n                   logx is one to one

7(n-2)=6n

7n-14=6n

n=14

Guest Feb 28, 2017

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