MATH 110 Sample 02 Exam 1

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1. Evaluate: log 4 64 [A] 3 [B] 1 12 [C] 12 [D] 1 3 2. Given log 5 = T and log 2 = U, find log. [A] T + U [B] TU [C] T + U [D] TU. 3. Find x if e x 52 8 =, and ou are given ln 8 = 2.0794. [A] 0.3999 [B] 0.1923 [C] 2.5007 [D] 0.6500 1 4. Evaluate: log 8 64 F I HG K J [A] 2 [B] 3 [C] 2 [D] 3

Graph: 5. f x 1 3 x 1 ( ) = F H G I K J 3 [A] [B] x x [C] [D] none of these x

Graph: 6. f(x) = F 1 HG I 2K J x [A] [B] x x [C] [D] x x

7. The amount of mone A accrued at the end of n ears when a certain amount P is invested at a compound annual rate r is given b A= P( 1 + r) n. If a person invests $180 at 6% interest compounded annuall, find the approximate amount obtained at the end of 15 ears. [A] $431 [B] $1260 [C] $18,900 [D] $207,526 8. Write as a single logarithm: 9 log x + 2 log [A] log b x b18xg [B] log b F 9 I HG 2 b b KJ [C] log b x F HG 9 x c h 9 2 [D] log b 2 I KJ 9. Write the equation 3 2 = 9 in logarithmic form. [A] log 9 3= 2 [B] log 1 9= 3 [C] log 2 9= 3 [D] log 3 9= 2 2

. Find ln 265. Round our answer to four decimal places. Solve: 2 2 11. x + = 4 2 2 x + = 1 9 49 [A] {(0, 7), (0, 7 )} [B] [C] {(0, 3), (0, 3 )} [D] {( 3, 0), ( 3, 0)} 12. 2 2 x + = 36 x+ = 6 [A] {(0, 6), (6, 0)} [B] {(6, 6), ( 6, 6)} [C] {(0, 6), ( 6, 0)} [D] {(0, 0), (6, 6)}

Solve: 3 2 13. x + x 25x 25 = 0 [A] 1, 5, 5 [B] 1, 5 [C] 5, 1, 5 [D] 1, 25 14. Graph the sstem: = ( x + 2) 2 = 2x + 4 [A] [B] x x [C] [D] x x

15. Solve the sstem: 2 2 x + = 63 x 3 = 27 2 2 [A] (3 6, 3), (3 6, 3) ( 3 6, 3), ( 3 6, 3) [C] (1, 62), (1, 62) ( 1, 62), ( 1, 62) [B] (3 6, 3), (3 6, 3) [D] ( 3 6, 3), ( 3 6, 3) Solve: 16. 2 25 / a 5a 15 / + 2 = 0 2 17. 2x + x 1 = 0

Solve: 18. x+ 34 = x+ 14 2 2 19. Solve the sstem: x + = 49 = 2x + 2

20. Which function matches the graph? 5 5 5 x 5 5 2 [A] f(x) = x + x + x [B] f(x) = 2x 4 + x 2 1 5 3 [C] f(x) = 2x + x 1 [D] f(x) = 2x 4 x 2 + 1 6 4 3 2 21. Use snthetic division to find f ( 5) if f ( x) = 2x 49x + 3x 14x + 5x+ 8.

22. Use the Intermediate Value Theorem to determine which interval contains an x-intercept of the function. f(x) = 8x 5 + 5x 4 6x 3 x 2 2x 2 [A] [ 2, 1] [B] none of these [C] [ 9, 8] [D] [3, 4] 6 5 3 2 23. Use the Remainder Theorem to find P( 4) if Px ( ) = x + 3x + x 4x 27. Also find the quotient polnomial that leads to the remainder. 5 4 3 2 [A] 852; x + 7x + 4x + 17x 64x 224 5 4 3 2 [B] 869; x x + 4x 15x + 56x 224 5 4 3 2 [C] 852; x x + 4x 15x + 56x 224 5 4 3 2 [D] 869; x + 7x + 4x + 17x 64x 224

24. Use Descartes Rule of Signs to determine how man positive and how man negative real zeros the polnomial functions ma have. Do not attempt to find the zeros. 6 5 4 3 2 f ( x) = x 4x + x 2x 2x + x+ 5 [A] 5, 3, or 1 positive real zeros; 2 or no negative real zeros [B] 4, 2, or no positive real zeros; 3 or 1 negative real zeros [C] 2 or no positive real zeros; 4, 2, or no negative real zeros [D] 4, 2, or no positive real zeros; 2 or no negative real zeros

25. Graph: f(x) = x + 4 x + 3 [A] [B] x x [C] [D] x x

[1] [2] [3] [4] [5] [6] [7] [8] [9] [] [11] [12] [13] [14] [15]

[16] [17] [18] [19] [20] [21] [22] [23] [24] [25]

[1] [A] [2] [C] [3] [A] [4] [A] [5] [B] [6] [C] [7] [A] [8] [C] [9] [D] [] 5.5797

[11] [B] [12] [A] [13] [A] [14] [A] [15] [A] [16] RST 32, 1 32 UVW [17] 1, 1 2 [18] 9 [19] F HG 4 + 5 241, 2 2 5 I F HG 241 KJ, 4 241, 2 + 2 5 5 241 I KJ

[20] [C] [21] 117 [22] [A] [23] [B] [24] [D] [25] [A]