Problem

Given \(m~=~\log_b(M)\) and \(n~=~\log_b(N)\), derive \(\log_b(MN)~=~\log_b(M)~+~\log_b(N)\) with a complete exponent chain and state all base and argument conditions.

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What this problem is really about

The product law is the equal-base exponent rule viewed through inverse functions. Translate each logarithm statement into an exponential equation, multiply those equations, and add the exponents on the shared base. Converting the product back to logarithmic form completes the chain, provided the base is valid and both original arguments are positive.

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Candidate analysis: M = bm and N = bn, so MN = bm⁺ⁿ. Therefore logb(MN) = m + n = logb(M) + logb(N), for positive M and N and valid base b.

Candidate analysis: M = bm and N = bⁿ, so MN = bm⁺ⁿ. Therefore logb(MN) = m + n = logb(M) + logb(N), for nonzero M and N and valid base b.

Candidate analysis: M = bm and N = bⁿ, so MN = bmⁿ. Therefore logb(MN) = mn = logb(M) + logb(N), for positive M and N and valid base b.

Candidate analysis: M = bm and N = bⁿ, so MN = bm⁺ⁿ. Therefore logb(MN) = m + n = logb(M) + logb(N), for positive M and N and valid base b.

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Given \(m~=~\log_b(M)\) and \(n~=~\log_b(N)\), derive \(\log_b(MN)~=~\log_b(M)~+~\log_b(N)\) with a complete exponent chain and state all base and argument conditions.

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