You have learned various rules for manipulating and simplifying expressions with exponents, such as the rule that says that x^{3} × x^{5} equals x^{8} because you can add the exponents. There are similar rules for logarithms.
Log Rules:
1) log_{b}(mn) = log_{b}(m) + log_{b}(n)
2) log_{b}(^{m}/_{n}) = log_{b}(m) – log_{b}(n)
3) log_{b}(m^{n}) = n · log_{b}(m)
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In less formal terms, the log rules might be expressed as:
1) Multiplication inside the log can be turned into addition outside the log, and vice versa.
2) Division inside the log can be turned into subtraction outside the log, and vice versa.
3) An exponent on everything inside a log can be moved out front as a multiplier, and vice versa.
Warning: Just as when you're dealing with exponents, the above rules work only if the bases are the same. For instance, the expression "log_{d}(m) + log_{b}(n)" cannot be simplified, because the bases (the "d" and the "b") are not the same, just as x^{2} × y^{3} cannot be simplified because the bases (the x and y) are not the same.
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Log rules can be used to simplify (or, more correctly, to "condense") expressions, to "expand" expressions, or to solve for values. We'll start with expansion.
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When the instructions say to "expand", they mean that they've given me one log expression with lots of stuff inside it, and they want me to use the log rules to take the log apart into many separate log terms, each with only one thing inside its particular log. That is, they've given me one log with a complicated argument, and they want me to convert this to many logs, each with a simple argument.
In this case, I have a "2x" inside the log. Since "2x" is multiplication, I can take this expression apart, according to the first of the log rules above, and turn it into an addition outside the log:
log_{3}(2x) = log_{3}(2) + log_{3}(x)
Then the answer they are looking for is:
log_{3}(2) + log_{3}(x)
Note: Do not try to evaluate "log_{3}(2)" in your calculator. While you would be correct in saying that "log_{3}(2)" is just a number (and we'll be seeing later how to rearrange this expression into something that you can evaluate in your calculator), what they're actually looking for here is the "exact" form of the log, as shown above, and not a decimal approximation from your calculator.
If you give "the answer" as being the decimal approximation, you should expect to lose points.
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I have division inside the log. According to the second of the log rules above, this can be split apart as subtraction outside the log, so:
log_{4}(^{ 16}/_{x}) = log_{4}(16) – log_{4}(x)
The first term on the right-hand side of the above equation can be simplified to an exact value, by applying the basic definition of what a logarithm is. In this case, I'm using the fact that the power required on 4 to create 16 is 2; in other words, since 4^{2} = 16, then:
log_{4}(16) = 2
Then the original expression expands fully as:
log_{4}(^{ 16}/_{x} ) = 2 – log_{4}(x)
Always remember to take the time to check to see if any of the terms in your expansion (such as the log_{4}(16) above) can be simplified.
The exponent inside the log can be taken out front as a multiplier:
log_{5}(x^{3}) = 3 · log_{5}(x) = 3log_{5}(x)
The examples above are very simple uses of the log rules, as applied to the expansion of log expressions. On the next page, we'll take a look at the sort of exercises you'll be seeing in your homework and on the next test.
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