+0  
 
0
147
3
avatar

what is the answer to:

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

Guest Feb 28, 2017
 #1
avatar
0

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
avatar+87301 
0

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
avatar
0

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

8 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.