Solve by equating exponents
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1 EXAMPLE 1 x Solve 4 = 1 Solve by equating exponents x 3 SOLUTION x 1 4 = x x 3 1 ( ) = ( ) x x + 3 = x = x + 3 x = 1 x 3 Write original equation. Rewrite 4 and base. 1 as powers with Power of a power property Property of equality for exponential equations Solve for x. ANSWER The solution is 1.
2 EXAMPLE 1 Solve by equating exponents Check: Check the solution by substituting it into the original equation. 1? 1 4 = 1 3 Substitute 1 for x.? 1 4 = 4 = 4 Simplify. Solution checks.
3 GUIDED PRACTICE for Example 1 Solve the equation. x x = 7 x 9 = 7 SOLUTION x x 1 9 = 7 x x 1 9 = (9 3) x x 1 3 (3 ) = (3 ) 4x 3x 3 3 = 3 4x = 3x 3 Write original equation. Rewrite 9and 7 as powers with base 3. Power of a power property Property of equality for exponential equations
4 GUIDED PRACTICE for Example 1 4x 3x = 3 Property of equality for exponential equations = 3 Solve for x. ANSWER The solution is 3.
5 GUIDED PRACTICE for Example 1 Solve the equation. 7x + 1 3x. 100 = 1000 SOLUTION 7x + 1 3x 100 = 1000 Write original equation. 7x+1 3x (10 ) = (10 ) 14x+ (10 ) = (10 ) 9x 6 3 Rewrite 100and 1000 as powers with base 10. Power of a power property 14x + = 9x 6 Property of equality for exponential equations
6 GUIDED PRACTICE for Example 1 14x 9x = 6 ANSWER 5x = 8 x = 8 5 The solution is 8. 5 Property of equality for exponential equations Property of equality for exponential equations Solve for x.
7 GUIDED PRACTICE for Example 1 Solve the equation. 3 x = 3 5x 6 SOLUTION 3 x 1 81 = 3 5x 6 Write original equation. 4 3 x 1 5x 6 (3 ) = (3 ) 1 4x 5x+6 3 = 3 Rewrite 81 and 1/3 as powers with base 3. Power of a power property 1 4x = 5x+6 Property of equality for exponential equations
8 GUIDED PRACTICE for Example 1 4x + 5x = 6 1 Property of equality for exponential equations x = 6 Solve for x. ANSWER The solution is 6.
9 EXAMPLE Take a logarithm of each side x Solve 4 = 11. SOLUTION x 4 = 11 x log 44 = log 411 x = log 4 11 log 11 x = log 4 x 1.73 Write original equation. Take log of each side. 4 x log b b = x Change-of-base formula Use a calculator. ANSWER The solution is about Check this in the original equation.
10 EXAMPLE 3 Use an exponential model Cars You are driving on a hot day when your car overheats and stops running. It overheats at 80 F and can be driven again at 30 F. If r = and it is 80 F outside,how long (in minutes) do you have to wait until you can continue driving?
11 EXAMPLE 3 SOLUTION Use an exponential model T = ( T T )e R rt + T 30 = (80 80)e t = 00e t 0.75 = e t ln 0.75 = ln e t t 60 t R Newton s law of cooling Substitute for T, T, T, and r. R Subtract 80 from each side. Divide each side by 00. Take natural log of each side. x In e = log e = x e x Divide each side by
12 EXAMPLE 3 Use an exponential model ANSWER You have to wait about 60 minutes until you can continue driving.
13 GUIDED PRACTICE for Examples and 3 Solve the equation. x 4. = 5 SOLUTION x = 5 x Log = log 5 x = log 5 x = log 5 log Write original equation. Take log of each side. x log b b = x Change-of-base formula x.3 Use a calculator.
14 GUIDED PRACTICE for Examples and 3 ANSWER The solution is about.3. Check this in the original equation.
15 GUIDED PRACTICE for Examples and 3 Solve the equation. 9x 5. 7 = 15 SOLUTION 9x 7 = 15 9x Log 77 = log 715 9x = log 7 15 Write original equation. Take log of each side. 7 x log b b = x 9x = log 15 log 7 Change-of-base formula x Use a calculator.
16 GUIDED PRACTICE for Examples and 3 ANSWER The solution is about Check this in the original equation.
17 GUIDED PRACTICE for Examples and 3 Solve the equation. 0.3x 6. 4e 7 = 13 SOLUTION 0.3x 4e 7 = x 4e = x 4e = 0 0.3x 0 e = 4 = 5 0.3x log e = log 5 e e
18 GUIDED PRACTICE for Examples and 3 x 0.3x = In 5 In e = log e = x e x 0.3x = Divide each side by 0.3x x = 0.3 = ANSWER The solution is about Check this in the original equation.
19 EXAMPLE 4 Solve a logarithmic equation Solve log (4x 7) = log (x + 5). 5 5 SOLUTION log (4x 7) = log (x + 5) x 7 = x x 7 = 5 3x = 1 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. ANSWER The solution is 4.
20 EXAMPLE 4 Solve a logarithmic equation Check: Check the solution by substituting it into the original equation. log 5 (4x 7) = log 5 (x 5) log? 5 (4 4 7) = log 5 (4 + 5) log 5 9 = log 5 9 Write original equation. Substitute 4 for x. Solution checks.
21 EXAMPLE 5 Exponentiate each side of an equation Solve log 4 (5x 1)= 3 SOLUTION log 4 (5x 1)= 3 4 log 4 (5x 1) = 4 3 5x 1 = 64 5x = 65 x = 13 Write original equation. Exponentiate each side using base 4. b log x b = x Add 1 to each side. Divide each side by 5. ANSWER The solution is 13.
22 EXAMPLE 5 Exponentiate each side of an equation Check: log 4 (5x 1) = log 4 (5 13 1) = log Because 4 = 64, log 4 64= 3.
23 EXAMPLE 6 Standardized Test Practice SOLUTION log x + log (x 5) = log [x(x 5)] = log [x(x 5)] 10 = 10 x(x 5) = 100 Write original equation. Product property of logarithms Exponentiate each side using base 10. Distributive property
24 EXAMPLE 6 Standardized Test Practice x 10x = 100 x 10x 100 = 0 x 5x 50 = 0 (x 10)(x + 5) = 0 x = 10 or x = 5 b log x b = x Write in standard form. Divide each side by. Factor. Zero product property Check: Check the apparent solutions 10 and 5 using algebra or a graph. Algebra: Substitute 10 and 5 for x in the original equation.
25 EXAMPLE 6 Standardized Test Practice log x + log (x 5) = log ( 10) + log (10 5) = log 0 + log 5 = log 100 = = log x + log (x 5) = log [( 5)] + log ( 5 5) = log ( 10) + log ( 10) = Because log ( 10) is not defined, 5 is not a solution. So, 10 is a solution.
26 EXAMPLE 6 Standardized Test Practice Graph: Graph y = log x + log (x 5) and y = in the same coordinate plane. The graphs intersect only once, when x = 10. So, 10 is the only solution. ANSWER The correct answer is C.
27 GUIDED PRACTICE for Examples 4, 5 and 6 Solve the equation. Check for extraneous solutions. 7. ln (7x 4) = ln (x + 11) SOLUTION ln (7x 4) = ln (x + 11) 7x 4 = x x x = 11 4 Write original equation. Property of equality for logarithmic equations ANSWER 5x = 15 x = 3 The solution is 3. Divide each side by 5.
28 GUIDED PRACTICE for Examples 4, 5 and 6 Solve the equation. Check for extraneous solutions. 8. log (x 6) = 5 SOLUTION log (x 6) = 5 Write original equation. log (x 6) = ANSWER 5 x 6 = 3 x = x = 38 The solution is 38. Exponentiate each side using base. b log x b = x Add 6 to each side.
29 GUIDED PRACTICE for Examples 4, 5 and 6 Solve the equation. Check for extraneous solutions. 9. log 5x + log (x 1) = SOLUTION log 5x + log (x 5) = log [5x(x 1)] = log [5x(x 1)] 10 = 10 5x(x 1) = 100 Write original equation. Product property of logarithms Exponentiate each side using base 10. Distributive property
30 GUIDED PRACTICE for Examples 4, 5 and x x = 5 x x = 0 x x 0 = 0 x 5x + 4x 0 = 0 x(x 5)+4(x 5) = 0 (x 5)(x +4) = 0 b log x b Factor. = x x = 5 or x = 4 Zero product property Check: Check the apparent solutions 5 and 4 using algebra or a graph.
31 GUIDED PRACTICE for Examples 4, 5 and 6 Algebra: Substitute 4 and 5 for x in the original equation. log 5x + log (x 1) = log5( 4 ) + log ( 4 1) = log 0 + log 5 = log 100 = log 5x + log (5x 1) = log [5(5)] + log (5(5) 1) = log 5 + log 4 = log 600 = ANSWER = So, 4 is a solution..778 =
32 GUIDED PRACTICE for Examples 4, 5 and 6 Solve the equation. Check for extraneous solutions. 10. log 4 (x + 1) + log 4 x =3 SOLUTION log 4 (x + 1) + log 4 x =3 log 4 [(x + 1) x] = 3 log 4 (x + 1 x) = 3 log (x + 1 x) 4 4 = x + 1 x = 4 Exponentiate each side using base 4.
33 GUIDED PRACTICE for Examples 4, 5 and 6 x + 1 x = 64 x + 1x 64 = 0 x +16x 4x 64 = 0 x(x +4) 4(x +4) = 0 (x 4)(x +4) = 0 x = 4 or x = 4 Factor. Zero product property Check: Check the apparent solutions 4and 4 using algebra or a graph.
34 GUIDED PRACTICE for Examples 4, 5 and 6 Algebra: Substitute 4 and 4 for x in the original equation. log 4 (4 + 1) + log 44 = 3 log log 44 = 3 log = 3 log 416 = log 16 = log = log 4 ( 4 + 1) + log 4 4 = 3 log 48 + log 4 4 = 3 log 48 1= 3 log 48 = 4 log 8 log 4 = = 4 ANSWER So, 4 is a solution.
35 EXAMPLE 7 Use a logarithmic model Astronomy The apparent magnitude of a star is a measure of the brightness of the star as it appears to observers on Earth. The apparent magnitude M of the dimmest star that can be seen with a telescope is given by the function M = 5 log D + where D is the diameter (in millimeters) of the telescope s objective lens. If a telescope can reveal stars with a magnitude of 1, what is the diameter of its objective lens?
36 EXAMPLE 7 SOLUTION Use a logarithmic model M = 5 log D + 1 = 5 log D + 10 = 5 log D = log D Log D 10 = = D Write original equation. Substitute 1 for M. Subtract from each side. Divide each side by 5. Exponentiate each side using base 10. Simplify. ANSWER The diameter is 100 millimeters.
37 GUIDED PRACTICE for Example WHAT IF? Use the information from Example 7 to find the diameter of the objective lens of a telescope that can reveal stars with a magnitude of 7. SOLUTION M = 5 log D + 7 = 5 log D + Write original equation. Substitute 1 for M. 5 = 5 log D 1 = log D Subtract from each side. Divide each side by 5. 1 Log D 10 = = D ANSWER Simplify. The diameter is 10 millimeters.
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