![]() Note that since we can only take the logarithm of positive numbers, the domain of the left-hand side function is the set of all numbers satisfying both x + 13 > 0 and x − 2 > 0, or equivalently x > − 13 and x > 2. The solution to the problem is their intersection. Use the (reverse of the) product rule for logarithms:Įach side of the above equation is graphed to the right. Suppose we want to find the solutions to this equation: The product rule for logarithms requires that all the logarithms appearing in the rule be properly defined. As a simple illustration, observe that the domain of the function y = log 3 x 2 is x ≠ 0, while the domain of y = 2 log 3 x is x > 0. As we will see in the examples below, algebraic manipulations of expressions involving logarithms can easily lead to "solutions" which are not valid because of this domain restriction. Ĭaution: When solving equations involving logarithms, it is very important to keep in mind that the domain of a logarithm function is the positive numbers. So, there are two solutions: x = 2 4 = 16 and x = 2 − 4 = 1 16. Taking the square roots of both sides gives log 2 x = ± 4. Substituting this into log 4 x ⋅ log 2 x = 8 and cross-multiplying by 2, we get log 2 x ⋅ log 2 x = log 2 x 2 = 16. ![]() Next, using the change of base rule, we have log 4 x = log 2 x log 2 4 = log 2 x 2. Solution: First, note that by the power rule log 2 x 2 = 2 log 2 x, so the original equation reduces to log 4 x ⋅ log 2 x = 8. If the equation can be manipulated into the form log b x = y (that is, involving just a single logarithm) then x = b y. If an equation with logarithms can be solved using algebraic techniques, then those techniques will generally involve the product, quotient, and power rules of logarithms-applied in either direction-as well as examining the problem for common bases.
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