In this section we will be working with Properties of Logarithms in an attempt to take equations with more than one logarithm and condense them down into just one single logarithm. Properties of Logarithms: a. Product Rule: log π (π) + log π (π) = log π (π β π) When two or more logarithms with the same base are added, those logarithms can be condensed into one logarithm whose argument is the product of the original arguments (π β π in the example above) Order is not important when multiplying or adding, so changing the order of the factors in an argument or changing the order of the terms being added together does not change the answer. b. Quotient Rule: π log π (π) β log π (π) = log π ( ) π When two or more logarithms with the same base are subtracted, those logarithms can be condensed into one logarithm whose argument is the quotient of the original arguments π ( ππ π‘βπ ππ₯πππππ ππππ£π) π Order is important when dividing and subtracting; the numerator is always the argument of the first log term listed (or the argument of the term listed first is always the numerator). c. Power Rule: π β log π (π) = log π (ππ ) A factor times a logarithm can be re-written as the argument of the logarithm raised to the power of that factor There is no Property of Logarithms that applies when the argument is a sum or difference. log c (π + π) β log c (π) + log c (π) log c (π β π) β log c (π) β log c (π) If an equation has more than one logarithm on either side of the equation, use the Properties of Logarithms to simplify as much as possible, then solve by converting to exponential form. REMEMBER THAT YOU MUST CHECK YOUR ANSWERS WHEN SOLVING LOG EQUATIONS TO VERIFY THAT THEY MAKE THE ORIGINAL ARGUMENTS POSITIVE. Example 1: Solve each of the following logarithmic equations and CHECK YOUR SOLUTIONS. a. 2 = log 3 (π₯) β log 3 (π₯ β 1) b. log 2 (π₯ + 7) + log 2 (π₯ ) = 3 c. 2 ln(π₯ ) β ln(5π₯ β 6) = 0 d. log(1 β π₯ ) = 1 β log(π₯ β 1) 1 e. 2 log(π₯ + 1) = log(1 β π₯ ) Example 2: Solve each of the following logarithmic equations and CHECK YOUR SOLUTIONS. a. log 5 (10) β log 5 (50) = log 5 (0.5) + log 5 (4π₯ + 6) b. log 4 (3π₯ + 2) = log 4 (5) + log 4 (3) c. ln(π₯ β 2) = 1 β ln(π₯) (hint: when you get a quadratic equation, solve by completing the square rather than using the quadratic formula) d. log(π₯ + 2) β log(π₯) = 2 log(4) e. log 3 (π₯ + 3) + log 3 (π₯ + 5) = 1 f. log(4 β 5π₯ ) β log(10 β π₯ 2 ) = 0 g. ln(π₯ ) + ln(6π₯ β 5) = 0 h. 2 log 8 (π₯) = 3 log 8 (4) Remember that there is no Property of Logarithms that applies when the argument is a sum or difference. log c (π + π) β log c (π) + log c (π) log c (π β π) β log c (π) β log c (π) 5 Therefore equations such as ln(π₯ 2 + 1) = 2 or log (π₯2 β 4) = 0 CANNOT be simplified using any Properties of Logarithms first before solving. Instead, these types of equations would simply be solved by converting to exponential form. Example 3: Solve each of the following logarithmic equations and CHECK YOUR SOLUTIONS. a. ln(π₯ 2 + 1) = 2 b. ln(π₯ 2 + 1) = β1 5 c. log (π₯ 2 β 4) = 0 Answers to Examples: 9 1a. π₯ = 8 ; 1b. π₯ = 1 ; 1c. π₯ = 2; π₯ = 3 ; 1d. ππ πππΏπππΌππ ; 1e. π₯ = 0 ; 7 13 2 2a. π₯ = β 5 ; 2b. π₯ = 3 ; 2c. π₯ = 1 + βπ + 1 ; 2d. π₯ = 15 ; 2e. π₯ = β2 ; 2f. π₯ = β1 ; 2g. π₯ = 1 ; 2h. π₯ = 8 ; 3 3a. π₯ = ±βπ2 β 1 ; 3b. ππ πππΏπππΌππ ; 3c. π₯ = ± 2;
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