EXAMPLE 4 Solve a logarithmic equation Solve log (4x – 7) = log (x + 5). 5 5 log (4x – 7) = log (x + 5). 5 5 4x – 7 = x + 5 3 x – 7 = 5 3x = 12 x = 4 Write.

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Presentation transcript:

EXAMPLE 4 Solve a logarithmic equation Solve log (4x – 7) = log (x + 5). 5 5 log (4x – 7) = log (x + 5) x – 7 = x x – 7 = 5 3x = 12 x = 4 Write original equation. Property of equality for logarithmic equations Subtract x from each side. Add 7 to each side. Divide each side by 3. The solution is 4. ANSWER SOLUTION

EXAMPLE 4 Solve a logarithmic equation Check: Check the solution by substituting it into the original equation. Write original equation. Substitute 4 for x. Solution checks. (4x – 7) = (x – 5) log 5 5 (4 4 – 7) = (4 + 5) ? log = 9 log 5 5

EXAMPLE 5 Exponentiate each side of an equation 5x – 1 = 64 5x = 65 x = 13 SOLUTION Write original equation. Exponentiate each side using base 4. Add 1 to each side. Divide each side by 5. Solve (5x – 1)= 3 log 4 4 log 4 (5x – 1) = 4 3 (5x – 1)=(5x – 1)= 3 log 4 b = x log b x The solution is 13. ANSWER

EXAMPLE 5 Exponentiate each side of an equation log 4 (5x – 1) = (5 13 – 1) = 64 log 4 4 Check: log 4 Because 4 = 64, 64= 3. 3

EXAMPLE 6 Standardized Test Practice SOLUTION log 2x + log (x – 5) = 2 log [2x(x – 5)] = 2 2x(x – 5) = 100 Write original equation. Product property of logarithms Exponentiate each side using base 10. Distributive property 10 = 10 log 2 [2x(x – 5)]

EXAMPLE 6 Standardized Test Practice x – 5x – 50 = 0 2 (x – 10)(x + 5) = 0 x = 10 or x = – 5 Check: Check the apparent solutions 10 and – 5 using algebra or a graph. Write in standard form. Divide each side by 2. Factor. 2x – 10x – 100 = 0 2 2x – 10x = Zero product property b = x log b x Algebra: Substitute 10 and 25 for x in the original equation.

EXAMPLE 6 log 2x + log (x – 5) = 2 Standardized Test Practice log 20 + log 5 = 2 log 100 = 2 2 = 2 log 2x + log (x – 5) = 2 log [2(–5)] + log (–5 – 5) = 2 log (–10) + log (–10) = 2 Because log (–10) is not defined, –5 is not a solution. So, 10 is a solution. log (2 10) + log (10 – 5) = 2

EXAMPLE 6 Standardized Test Practice Graph: Graph y = log 2x + log (x – 5) and y = 2 in the same coordinate plane. The graphs intersect only once, when x = 10. So, 10 is the only solution. The correct answer is C. ANSWER

GUIDED PRACTICE for Examples 4, 5 and 6 Solve the equation. Check for extraneous solutions. 7. ln (7x – 4) = ln (2x + 11) SOLUTION Write original equation. ln (7x – 4) = ln (2x + 11) 7x – 4 = 2x x – 2x = 11 – 4 5x = 15 x = 3 The solution is 3. ANSWER Property of equality for logarithmic equations Divide each side by 5.

GUIDED PRACTICE for Examples 4, 5 and 6 Solve the equation. Check for extraneous solutions. 8. log (x – 6) = 5 2 SOLUTION log (x – 6) = 5 2 Write original equation. 2log (x – 6) = x – 6 = 32 x = x = 38 The solution is 38. ANSWER Exponentiate each side using base 2. b = x log b x Add 6 to each side.

GUIDED PRACTICE for Examples 4, 5 and 6 9. log 5x + log (x – 1) = 2 Solve the equation. Check for extraneous solutions. SOLUTION log 5x + log (x – 5) = 2 log [5x(x – 1)] = 2 5x(x – 1) = 100 Write original equation. Product property of logarithms Exponentiate each side using base 10. Distributive property 10 = 10 log 2 [5x(x – 1)]

GUIDED PRACTICE for Examples 4, 5 and 6 x – 5x + 4x – 20 = 0 2 (x – 5)(x +4) = 0 x = 5 or x = – 4 Factor. x – x – 20 = 0 2 x – x = 20 2 Zero product property b = x log b x x – x = x(x – 5)+4(x – 5) = 0 Check: Check the apparent solutions 5 and – 4 using algebra or a graph.

GUIDED PRACTICE for Examples 4, 5 and 6 Algebra: Substitute – 4 and 5 for x in the original equation. log 5x + log (x – 1) = 2 log –20 + log –5 = 2 log 100 = 2 2 = 2 log 5x + log (5x – 1) = 2 log [5(5)] + log (5(5) –1) = 2 log 25 + log 24 = 2 log5(– 4 ) + log (– 4 –1) = 2 log 600 = = 2 So, – 4 is a solution. ANSWER

GUIDED PRACTICE for Examples 4, 5 and 6 Solve the equation. Check for extraneous solutions. SOLUTION 10. log (x + 12) + log x =3 4 4 log (x + 12) + log x = log [(x + 12) x] = 3 4 log (x + 12 x) = 3 2 x + 12 x = Exponentiate each side using base 4. 4 log (x + 12 x) 2 4 = 4 3

GUIDED PRACTICE for Examples 4, 5 and 6 (x – 4)(x +4) = 0 x = 4 or x = – 4 x + 12x – 64 = 0 2 x(x +4) – 4(x +4) = 0 x + 12 x = 64 2 x +16x – 4x – 64 = 0 2 Factor. Zero product property Check: Check the apparent solutions 4 and – 4 using algebra or a graph.

GUIDED PRACTICE for Examples 4, 5 and 6 Algebra: Substitute 4 and – 4 for x in the original equation. So, 4 is a solution.ANSWER log (4 + 12) + log 4 = log = 3 4 log log 4 = 3 4 log 16 = 2 log 16 log 4 = = 2 log (– ) + log – 4 = log 8 – 1= 3 4 log – 4 = 3 log 8 log 4 = 4 4 log 8 = = 4