WARMUP Lesson 7.6, For use with pages 515-522 1. Write log3(2x – 7) = 4 in exponential form. ANSWER 34 = 2x – 7 Write 8x = 30 in logarithmic form. Hint: reverse loop swoop doop ANSWER log8 30 = x Solve for x. ANSWER 3 2 3. 100x = 1000 ANSWER 1 125 4. log5x = –3 x2 – 7x – 60 = 0 Hint: Use the quadratic formula. ANSWER –5, 12

7.6 Notes - Solve Exponential and Log Equations https://www.youtube.com/watch?v=XjzBpJ1XYwc

Objective - To solve exponential and logarithmic Rewrite each base as a power of 2. Distribute. Powers equal to each other. Objective - To solve exponential and logarithmic equations.

Take the log3 of each side.

Take the log5 of each side.

If logb x = logb y if and only if x = y.

Loop Swoop Doop

If x = y, then bx = by. You do this one You probably did loop-swoop-doop, but here is another approach.

Check! Remember, you can’t take the log of a negative! 10 Check! Remember, you can’t take the log of a negative!

Solve by equating exponents EXAMPLE 1 Solve by equating exponents Solve 4 = x 1 2 x – 3 SOLUTION 4 = x 1 2 x – 3 Write original equation. Rewrite 4 and as powers with base 2. 1 2 (2 ) = (2 ) 2 x – 3 x – 1 2 = 2 2 x – x + 3 Power of a power property 2x = –x + 3 Property of equality for exponential equations x = 1 Solve for x. The solution is 1. ANSWER

Solve by equating exponents EXAMPLE 1 Solve by equating exponents Check: Check the solution by substituting it into the original equation. 1 – 3 4 = 1 ? 2 Substitute 1 for x. 4 = ? 1 2 – 2 Simplify. 4 = 4 Solution checks.

GUIDED PRACTICE for Example 1 Solve the equation. 3. 81 = 1 3 3. 81 = 3 – x 1 3 5x – 6 1. 9 = 27 2x x – 1 SOLUTION –3 SOLUTION –6 2. 100 = 1000 7x + 1 3x – 2 – 8 5 SOLUTION

Take a logarithm of each side EXAMPLE 2 Take a logarithm of each side Solve 4 = 11. x SOLUTION 4 = 11 x Write original equation. log 4 = log 11 x 4 Take log of each side. 4 log 4 x = 11 log b = x b x x = log 11 log 4 Change-of-base formula x 1.73 Use a calculator. The solution is about 1.73. Check this in the original equation. ANSWER

EXAMPLE 3 Use an exponential model You are driving on a hot day when your car overheats and stops running. It overheats at 280°F and can be driven again at 230°F. If r = 0.0048 and it is 80°F outside,how long (in minutes) do you have to wait until you can continue driving? Cars

Use an exponential model EXAMPLE 3 Use an exponential model SOLUTION T = ( T – T )e + T ° R – rt Newton’s law of cooling 230 = (280 – 80)e + 80 –0.0048t Substitute for T, T , T , and r. ° R 150 = 200e –0.0048t Subtract 80 from each side. 0.75 = e –0.0048t Divide each side by 200. ln 0.75 = ln e –0.0048t Take natural log of each side. –0.2877 –0.0048t In e = log e = x e x 60 t Divide each side by –0.0048.

EXAMPLE 3 Use an exponential model You have to wait about 60 minutes until you can continue driving. ANSWER

GUIDED PRACTICE for Examples 2 and 3 Solve the equation. 4. 2 = 5 4. 2 = 5 x SOLUTION about 2.32 5. 7 = 15 9x SOLUTION about 0.155 6. 4e –7 = 13 –0.3x SOLUTION about –5.365

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

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

Exponentiate each side of an equation EXAMPLE 5 Exponentiate each side of an equation Solve (5x – 1)= 3 log 4 SOLUTION (5x – 1)= (5x – 1)= 3 log 4 Write original equation. 4log4(5x – 1) = 4 3 Exponentiate each side using base 4. b = x log b x 5x – 1 = 64 5x = 65 Add 1 to each side. x = 13 Divide each side by 5. The solution is 13. ANSWER

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

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

Standardized Test Practice EXAMPLE 6 Standardized Test Practice 2x – 10x = 100 2 b = x log b x 2x – 10x – 100 = 0 2 Write in standard form. x – 5x – 50 = 0 2 Divide each side by 2. (x – 10)(x + 5) = 0 Factor. x = 10 or x = – 5 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.

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

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

Solve the equation. Check for extraneous solutions. GUIDED PRACTICE for Examples 4, 5 and 6 Solve the equation. Check for extraneous solutions. 7. ln (7x – 4) = ln (2x + 11) 9. log 5x + log (x – 1) = 2 SOLUTION 5 SOLUTION 3 8. log (x – 6) = 5 2 10. log (x + 12) + log x =3 4 SOLUTION 4 SOLUTION 38

EXAMPLE 7 Use a logarithmic model 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 Astronomy M = 5 log D + 2 where D is the diameter (in millimeters) of the telescope’s objective lens. If a telescope can reveal stars with a magnitude of 12, what is the diameter of its objective lens?

Use a logarithmic model EXAMPLE 7 Use a logarithmic model SOLUTION M = 5 log D + 2 Write original equation. 12 = 5 log D + 2 Substitute 12 for M. 10 = 5 log D Subtract 2 from each side. 2 = log D Divide each side by 5. Exponentiate each side using base 10. 10 = 10 2 Log D 100 = D Simplify. The diameter is 100 millimeters. ANSWER

GUIDED PRACTICE for Example 7 11. 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 The diameter is 10 millimeters.

7.6 Assignment   Hint #3: Quadratic Formula

WARMUP Daily Homework Quiz For use after Lesson 7.6 Solve. ANSWER 6 5 1. 25x = 125 –x + 2 2. 8x = 5 ANSWER about 0.77 3. log7 (5x – 8) = log7 (2x + 19) ANSWER 9 4. log3 (5x + 1) = 4 16 ANSWER 5. log5 5x + log5 (x – 4) = 2 5 ANSWER Bonus. Boiling water has a temperature of 212° F. Water has a cooling rate of r = 0.042. Use the formula T = (T0 – TR)e –rt + TR to find the number of minutes t it will take for the water to cool to a temperature of 80°F if the room temperature is 72°F. ANSWER about 13 min

Chapter 7 Test Review Assignment p.541: 17-20 all, 24-34 all (16 Q’s) p.543: 10-12 all, 16-24 all, 28 (A = Pert) (13 Q’s) Do as much as you can without a calculator. Use your calculator to check (if not in the back)

Answers to the Review (p.541) 16 Q’s -3 -1/3 6.86mm, 5.39mm log83 + log8 x + log8 y ln10 + 3ln x + ln y log8 – 4 log y ln 3 + ln y – 5 ln x 28. ln 3 + ln y – 5 ln x 29. log7 384 30. ln(12/x2) 31. ln36 32. 2.153 33.7 34. 3.592

Answers to the Test (p.543) 13 Q’s 2 -5 16. ln(49/64) 17. log496 log(5x/9) 2.431 1.750 1.732 0.874 104 2 28. $3307.82