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

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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.

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