how do i work problem logMN=log + log
let log (M) + log(N) = x then 10 ^ [ log(M) + log (N) ] = 10 ^ x. so simplifying we get
10 ^ log(M) *10 ^ log(N) = M*N = 10 ^ x. further simplification we get
log(MN) = x = log(M) + log (N)
how do i work problem logMN=log + log
let log (M) + log(N) = x then 10 ^ [ log(M) + log (N) ] = 10 ^ x. so simplifying we get
10 ^ log(M) *10 ^ log(N) = M*N = 10 ^ x. further simplification we get
log(MN) = x = log(M) + log (N)
A simple derivation:
log_{b }MN = log_{b} (b^{logb M }b^{logb N}) = log_{b} b^{logb M + logb N} = log_{b }M + log_{b }N.
Nastassia ---
The other answers by Will and Robert are correct.
I would just add an example:
log 12 = log 4 + log 3 OR log 6 + log 2, or even log 9 + log (4/3). Try those on your caluclator.
If you have something like log 3x, it would be log 3 + log x.
If I understand the question, this is just a property of logs. The log of a product is the sum of the logs.
log (M * N) = log (M) + log (N)
logMN=logM + logN. This is the log property for production.