All courses Math III · F-LE.4.1 30 of 55
Prove simple logarithm laws.

Expand a logarithm of a product without losing its domain

Problem
Assume b>0, b!=1, and x>0. Expand logb(6x) completely with the product law. Enter the factors, real domain, expanded form, and whether the rewrite preserves that stated 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 the real-log conditions

\[b>0,~b\ne~1,~\text{and}~x>0\]

A real logarithm needs a valid base and a positive argument. The prompt makes each factor in the separated form positive.

02

Identify multiplicative factors

\[\text{factors}:~6~\text{and}~x\]

Only multiplication creates a product-law split. Any grouped sum or difference remains one argument.

03

Apply the product law completely

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

Give each multiplicative factor its own logarithm and keep the original log base on every term.

04

Compare domains

\[\text{original}~\text{and}~\text{expanded}~\text{real}~\text{domain}:~x>0\]

Under the stated positivity conditions, both sides are defined on the same domain, so the identity is domain-preserving.

05

Submit all fields

\[\text{factors}=6~\text{and}~x;~\text{real}~\text{domain}=x>0;~\text{expanded}~\text{form}=\log_b(6)+\log_b(x);~\text{equality}~\text{and}~\text{stated}~\text{domain}~\text{preserved}=\text{yes}\]

The response records structure, domain, transformation, and validity rather than giving an unsupported symbolic rewrite.

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Another way

  1. Condense logb(6)+logb(x) with the product law in reverse; it returns logb(6x) on x>0.

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Common mistake

Do not split a grouped sum or difference, and do not omit the positivity conditions required by the separated logarithms.