Problem

Given \(b~>~0\), \(b~\ne~1\), and \(M~>~0\), derive \(\log_b(M^r)~=~\text{rlog}_b(M)\) with a complete exponent-chain proof and state the domain condition.

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What this problem is really about

Derive the power law by translating the logarithm into an exponential statement. Raising that statement to a real power invokes the power-of-a-power rule, which multiplies the exponents; translating back produces the logarithmic multiplier. The positive-argument condition is essential because it keeps every real power and every logarithm in the chain defined.

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Candidate analysis: M = bm, so Mr = (bm)r = bm⁺r. Therefore logb(Mr) = m + r = rlogb(M), with M > 0.

Candidate analysis: M = bm, so Mr = (bm)r = bmr. Therefore logb(Mr) = mr = rlogb(M), with M ≠ 0.

Candidate analysis: M = bm, so Mr = (bm)r = bmr. Therefore logb(Mr) = mr = rlogb(M), with M > 0.

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Given \(b~>~0\), \(b~\ne~1\), and \(M~>~0\), derive \(\log_b(M^r)~=~\text{rlog}_b(M)\) with a complete exponent-chain proof and state the domain condition.

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