SOLUTION: Since a logarithm is an exponent, how do you think the log property logb(xy) = logb (x) + logb (y) is related to the exponent property (b^m)(b^n) = b^(m+n)?

Algebra ->  College  -> Linear Algebra -> SOLUTION: Since a logarithm is an exponent, how do you think the log property logb(xy) = logb (x) + logb (y) is related to the exponent property (b^m)(b^n) = b^(m+n)?      Log On


   

Question 6831: Since a logarithm is an exponent, how do you think the log property
logb(xy) = logb (x) + logb (y) is related to the exponent property
(b^m)(b^n) = b^(m+n)?
Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
First let +J+=+log%28b%2Cx%29+%2B+log%28b%2Cy%29+

So then +log%28b%2Cxy%29+=+J+. The equivalent exponent for that log equation is +b%5EJ+=+xy+.

Now, let's put back the logs in place of the J (un-substitute):

+b%5E%28log%28b%2Cx%29+%2B+log%28b%2Cy%29%29+=+xy+.

Remember the rule that says +b%5E%28m+%2B+n%29+=+b%5Em%2Ab%5En+

Applying that above exponential rule, +b%5Elog%28b%2Cx%29%2Ab%5Elog%28b%2Cy%29+=+xy+.

Remember the rule +b%5Elog%28b%2Cm%29+=+m+. In this case, +b%5Elog%28b%2Cx%29+=+x+ and +b%5Elog%28b%2Cy%29+=+y+. So we're all cool.