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Log Rules / Expanding Sections: Basic log rules, Expanding, Simplifying, Trick questions, ChangeofBase formula 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) 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 x and y are not the same). Expanding logarithms Log rules can be used to simplify expressions, to "expand" expressions, or to solve for values.
When they say to "expand", they mean that they've given you one log expression with lots of stuff inside it, and they want you to use the log rules to take the log apart into lots of separate logs, each with only one thing inside. That is, they've given you one log with a complicated argument, and they want you to convert this to many logs, each with a simple argument.
I have a "2x" inside the log. Since "2x" is multiplication, I can take this expression apart and turn it into an addition outside the log: log_{3}(2x) = log_{3}(2) + log_{3}(x) The answer they are looking for is: log_{3}(2) + log_{3}(x) 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, they're actually looking here for the "exact"
form of the log, as shown above, and not a decimal approximation from
your calculator.
I have division inside the log, which 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 righthand side of the above equation can be simplified to an exact value, by applying the basic definition of what a logarithm is: 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) Top  1  2  3  4  Return to Index Next >>



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