SOLUTION: Let M=B^a and N=B^c. Show why log(base b)MN = log(base b)M + log(base b)N.

SOLUTION: Let M=B^a and N=B^c. Show why log(base b)MN = log(base b)M + log(base b)N.

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Question 67685This question is from textbook Advanced mathematics
: Let M=B^a and N=B^c. Show why log(base b)MN = log(base b)M + log(base b)N. This question is from textbook Advanced mathematics

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Let M=B^a and N=B^c. Show why log(base b)MN = log(base b)M + log(base b)N.
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Log(base b) MN = log(base b)(b^a*b^c)
log(base b)b^(a+c) a+c
But a =log(base b)M
and c = log(base b)N
So, log(base b)MN= log(base b)M + log(base b)N
Cheers,
Stan H.
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