ALGEBRA II CHAPTER 6 PRACTICE TEST (SECTIONS 6.1 THROUGH 6.6) Name_____Key______________________________ Period______ Date____________ Learning Target #27: I can create, graph, and evaluate exponential functions. Learning Target #28: I can apply natural base e to exponential functions. Graph the equation. 1. π¦ = (2)π₯ 2. π¦ = 3π π₯ y 7 6 5 4 3 2 1 y (2,4) (1,2) (0,1) -4 -3 -2 -1 18 16 14 12 10 8 6 (1,8.15) 4 2 1 -1 x 2 3 4 -3 (0,3) -2 (-1,1.1) -1 -2 1 x 2 3 3. The number of cell phone subscribers π¦ (in millions) can be approximated by the model π¦ = 233(1.06)π‘ , where π‘ is the number of years since 2006. a. Tell whether the model represents exponential growth or exponential decay. Growth; b-value is bigger than 1 b. Identify the annual percent increase or decrease in the population. 6% increase c. Estimate when the number of subscribers will be 300,000,000. About 4 years 4. You deposit $6000 in an account that pays 4.25% annual interest. Find the balance after π ππ‘ 4 years when the interest is compounded semi-annually. π΄ = π (1 + π) . 0425 2(4) ) π΄ = 6000 (1 + = $7,099.17 2 Simplify the expression. 5. 8π 7 24π 9 1 3π 2 Algebra II Chapter 6 TestβLT #27-#33 6. (3π 2π₯ )4 34 π 8π₯ = 81π 8π₯ Practice Test Learning Target #29: I can evaluate and simplify logarithm expressions. 7. Rewrite log 6 36 = 2 in exponential form. 8. 62 = 36 1 Rewrite 8β2 = 64 in logarithmic form. log 8 Evaluate the logarithm. 9. log 9 81 10. 9? = 81 2 log 2 1 = β2 64 1 8 2? = β3 Find the inverse of the function. 11. π¦ = 7π₯ 12. 1 8 π¦ = log(4π₯) 10π₯ 4 log 7 π₯ Learning Target #30: I can graph logarithmic functions. Learning Target #31: I can perform transformations of exponential and logarithmic functions. Describe the transformation of π(π) represented by π(π). Then graph π(π). π (π₯ ) = 4π₯ ; π(π₯ ) = 4π₯β1 + 1 13. 14. 1 π₯ 1 π₯+2 π (π₯ ) = ( ) ; π ( π₯ ) = ( ) β1 2 2 Right 1 and Up 1 Left 2 and Down 1 y (1,2) (0,1.25) Algebra II Chapter 6 TestβLT #27-#33 y x (-4,3) (-3,1) (-2,0) x Practice Test Write a rule for π . Let the graph of g be a horizontal stretch by a factor of 2, followed by a translation 4 15. units right and 3 units down of the graph of π(π₯ ) = 4π₯ . 1 π(π₯ ) = 4(2)π₯β4 β 3 16. Let the graph of g be a translation 3 units right, followed by a reflection in the π₯-axis of the graph of π (π₯ ) = 9π₯ . π(π₯ ) = β9π₯β3 Graph the function. π (π₯ ) = log 4 (π₯ + 1) β 2 17. 9 y 8 7 6 5 4 3 2 1 Left 1 and Down 2 x -1 -1 1 2 3 4 5 6 7 8 9 -2 (3,-1) -3 (0,-2) -4 Use the Change of Base Formula to evaluate each expression. Show your work and round your answer to four decimal places. 1 log 4 64 18. 19. log 4 1024 log 64 =3 1 log 4 log (1024) = β5 log 4 Learning Target #32: I can apply properties of logarithms. Expand or condense the logarithmic expression. log 5 7π 20. 21. log 3 81 + 5 log 3 2 log 5 7 + log 5 πΆ 22. log 4 + 2 log 3 β log 9 log 4 β 32 9 Algebra II Chapter 6 TestβLT #27-#33 log 3 81 β 25 23. π₯ 3 log ( ) π€ 3 log π₯ β 3 log π€ Practice Test Learning Target #33: I can solve exponential and logarithmic equations and inequalities. Solve the equation/inequality. Round to the nearest hundredth when necessary. 24. 25. 2π 5π₯ = 26 53π₯β2 = 1252π₯+5 log 53π₯β2 = log 1252π₯+5 (3π₯ β 2) log 5 = (2π₯ + 5) log 125 3π₯ β 2 = (2π₯ + 5)(3) 3π₯ β 2 = 6π₯ + 15 β17 = 3π₯ 17 π₯=β 3 π 5π₯ = 13 ln π 5π₯ = ln 13 5π₯ = ln 13 π₯ β .51 26. 34π₯β5 < 8 27. log 34π₯β5 < log 8 (4π₯ β 5) log 3 < log 8 4π₯ β 5 < 1.89279 π₯ < 1.72 28. log 2 3π₯ = log 2 (4π₯ β 2) log 7 2(π₯ + 5) = 10 710 = 2π₯ + 10 π₯ = 141,237,619.5 29. 4 log 2 π₯ β log 2 5 = log 2 125 2 log 6 π₯ = 3 2 63 = π₯ π₯ β 3.30 3π₯ = 4π₯ β 2 π₯=2 30. log 7 2 + log 7 (π₯ + 5) = 10 31. ln π₯ + ln(π₯ β 1) = 6 π₯4 log 2 = log 2 125 5 4 π₯ = 125 5 π₯ 4 = 625 π₯=5 Algebra II Chapter 6 TestβLT #27-#33 Practice Test
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