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The Product Rule

Part of the Exponentials, Logs & Sequences section of Coddy's Algebra 2 journey. Lesson 18 of 46.

Take log⁡3(9⋅27)\log_{3}\left(9\cdot 27\right). Since 9=329 = {3}^{2} and 27=3327 = {3}^{3}, the product is 32⋅33{3}^{2}\cdot {3}^{3}, which is 35{3}^{5}: exponents add when powers multiply. So the log is 5, which is 2 plus 3, the log of 9 plus the log of 27. That is the product rule:

log⁡b(MN)=log⁡b(M)+log⁡b(N)\log_{b}\left(MN\right) = \log_{b}\left(M\right)+\log_{b}\left(N\right)

It holds for any positive M and N, for the same reason. Logs are exponents: if M is b to the power p and N is b to the power q, then MN is b to the power p+qp+q. So the log of the product is the sum of the logs.

The rule is most useful when one factor is a letter. For positive x, log⁡3(9x)\log_{3}\left(9x\right) becomes log⁡3(9)+log⁡3(x)\log_{3}\left(9\right)+\log_{3}\left(x\right), and log⁡3(9)\log_{3}\left(9\right) is 2, so it is 2+log⁡3(x)2+\log_{3}\left(x\right).

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Challenge

Rewrite log⁡2(8x)\log_{2}\left(8x\right) with the product rule. Type each line in the box under the problem: first the sum of two logs, then with the number's log worked out. You are done when it reads log⁡2(x)\log_{2}\left(x\right) plus a whole number.

Try it yourself

log⁡2(8x)\log_{2}\left(8x\right)

Rewrite as log⁡2(x)+3\log_{2}\left(x\right)+3

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