493 4.5 Exponential and Logarithmic Equations CAUTION Recall that the domain of y = loga x is 10, ∞2. For this reason, it is always necessary to check that proposed solutions of a logarithmic equation result in logarithms of positive numbers in the original equation. EXAMPLE 7 Solving a Logarithmic Equation Solve log2 313x - 721x - 424 = 3. Give exact value(s). SOLUTION log2 313x - 721x - 424 = 3 13x - 721x - 42 = 23 Write in exponential form. 3x2 - 19x + 28 = 8 Multiply. Apply the exponent. 3x2 - 19x + 20 = 0 Write in standard form. 13x - 421x - 52 = 0 Factor. 3x - 4 = 0 or x - 5 = 0 Zero-factor property x = 4 3 or x = 5 Solve for x. A check is necessary to be sure that the argument of the logarithm in the given equation is positive. In both cases, the product 13x - 721x - 42 leads to 8, and log2 8 = 3 is true. The solution set is E 4 3 , 5F. S Now Try Exercise 53. EXAMPLE 8 Solving a Logarithmic Equation Solve log 13x + 22 + log 1x - 12 = 1. Give exact value(s). SOLUTION log 13x + 22 + log 1x - 12 = 1 log10 313x + 221x - 124 = 1 log x = log10 x; product property 13x + 221x - 12 = 101 Write in exponential form. 3x2 - x - 2 = 10 Multiply; 101 = 10. 3x2 - x - 12 = 0 Write in standard form. x = -b {2b2 - 4ac 2a Quadratic formula x = -1-12 {21-122 - 41321-122 2132 Substitute a=3, b= -1, c = -12. The two proposed solutions are 1 - 2145 6 and 1 + 2145 6 . The first proposed solution, 1 - 2145 6 , is negative. Substituting it for x in log 1x - 12 results in a negative argument, which is not allowed. Therefore, this solution must be rejected. The second proposed solution, 1 + 2145 6 , is positive. Substituting it for x in log 13x + 22 results in a positive argument. Substituting it for x in log 1x - 12 also results in a positive argument. Both are necessary conditions. Therefore, the solution set is E1 + 2145 6 F . S Now Try Exercise 77.
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