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All courses Algebra II · F-LE.4.1 37 of 55
Prove simple logarithm laws.

Expand a logarithm of a product without losing its domain

Problem
Assume \(b~>~0\), \(b~\ne~1\), and \(x~>~0\). Expand \(\log_b(6x)\) using the product law and state the validity domain.
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Hint

Read the argument as the product of 6 and x.

For positive factors, logb(MN)=logb(M)+logb(N). Addition or subtraction inside one factor does not split.

Solution walkthrough

01

Check logarithm conditions

\[b>0,~b\ne~1;~\log_b(6x)~\text{requires}~6x>0~->~x>0\]

A real logarithm needs a positive argument and a positive base other than 1. Since 6 is positive, the maximal x-domain is x>0.

02

Apply the product law

\[\log_b(6x)=\log_b(6)+\log_b(x)\]

On x>0, both factors 6 and x are positive, so each separated logarithm is defined and the product law is domain-preserving.

03

State expansion and domain

\[\text{Candidate}~\text{analysis}:~\log_b(6)~+~\log_b(x),~\text{valid}~\text{for}~x~>~0.\]

The expansion is valid exactly on the original real-log domain under the stated base conditions.

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Another way

  1. Verify by exponentiating: if u=log base b of 6 and v=log base b of x, then bu⁺v=6x.

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Common mistake

Do not split a logarithm before checking that every separated argument is positive. Here the factor 6 is positive and x must be positive.

Solution walkthrough video