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Dubai

Notes On Logs And Exponents

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Published in: Mathematics
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Practice problems with Solutions

Mukul D / Dubai

6 years of teaching experience

Qualification: Bachelor's of Engineering

Teaches: Chemistry, Mathematics, SAT, ACT, Maths

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  1. Logs & Exponents 2008-14 with MS 1.[5 marks] 2 — logs (x + 7) = logx Solve the equation 3 2.[5 marks] Let f = . The graph offis transformed into the graph of the function g by a translation of 3 —2 , followed by a reflection in the x-axis. Find an expression for 9@), giving your answer as a single logarithm. 3.[6 marks] The first terms of an arithmetic sequence are logsæ log32X Find x if the sum of the first 20 terms of the sequence is equal to 100. 4.[5 marks] Solve the equation4Z 1 + 8. 5.[6 marks] Solve the following system of equations. logt+ly = 2 1 4 6.[5 marks] Considera = 10823 x log34 x log45 x . 7.[5 marks] log3132. Given that a e Z, find the value of a. Solve the equation 8t 1 — 63æ. Express your answer in terms of In 2 and In 3.
  2. Logs & Exponents 2008-14 MS 1.[5 marks] Markscheme = og3 MIMIAI Note: Award Ml for changing to single base, Ml for incorporating the 2 into a log and Al for a correct equation with maximum one log expression each side. 18XM1 — FAI [5 marks] Examiners report Some good solutions to this question and few candidates failed to earn marks on the question. Many were able to change the base of the logs, and many were able to deal with the 2, but ofthose who managed both, poor algebraic skills were often evident. Many students attempted to change the base into base 10, resulting in some complicated algebra, few of which managed to complete successfully. 2.[5 marks] Markscheme h(æ) — 3) — 2 In(x — 3) — — —h(x) = 2 — — 3) Ml Note: Award MI only if it is clear the effect of the reflection in the x-axis: the expression is correct OR there is a change of signs of the previous expression OR there's a graph or an explanation making it explicit = Ine2 — In(æ — 3) Ml -Ine Al [5 marks] Examiners report This question was well attempted but many candidates could have scored better had they written down all the steps to obtain the final expression. In some cases, as the final expression was incorrect and the middle steps were missing, candidates scored just 1 mark. That could be a consequence of a small mistake, but the lack of working prevented them from scoring at least all method marks. Some candidates performed the transformations well but were not able to use logarithms properties to transform the answer and give it as a single logarithm. 3.[6 marks] Markscheme METHOD 1 log2 (MI) log28 low (Ml) Note: Award this MI for a correct change of base anywhere in the question. log2X (A I) 2
  3. — log2Z 100 — — log2C log.2x = 4 2 METHOD 2 2} 2 19 x —2— log2 22•r — + 8 Ml 4 16M 20th term = 20 100= 2 20 100 = log239 A I Ml log2 239 MI(AI) Note: Award this MI for a correct change of base anywhere in the question. 400 100 log2C log.2x = 4 2 METHOD 3 log2r logsx loge r 1 log. log,32 log2r low logzx Note: Award this MI for a correct change of base anywhere in the question. = (1+3+54...) Al — (2 + 38)) log.ar 2 (Ml) (Al) 4 16 Al 10428 x log2128 log.2 100 = loga log2a: = 4 = 2 [6 marks] Examiners report = 16M There were plenty of good answers to this question. Those who realised they needed to make each log have the same base (and a great variety of bases were chosen) managed the question successfully. 4.[5 marks] Markscheme 22—2 = + 8 (Ml) (Al) 4 x -32 -0M Notes: Do not award final Al if more than 1 solution is given. [5 marks] Examiners report 3
  4. Very few candidates knew how to solve this equation. A significant number guessed the answer using trial and error after failed attempts to solve it. A number of misconceptions were identified involving properties of logarithms and exponentials. 5.[6 marks] Markscheme logy +1 c = so (r + = YAI EITHER —1 = Ml C — —l, not possible RI OR Ml attempt to solve or graph of LHS MI 7.27 MAI [6 marks] Examiners report This question was well answered by a significant number of candidates. There was evidence of good understanding of logarithms. The algebra required to solve the problem did not intimidate candidates and the vast majority noticed the necessity of technology to solve the final equation. Not all candidates recognized the extraneous solution and there were situations where a rounded value of was used to calculate the value of Y leading to an incorrect solution. 6.[5 marks] Markscheme log 3 log 4 log 2 X log 3 log 32 log 2 Al 5 log2 (Ml) hencea 5 log 32 MIAI Note: Accept the above if done in a specific base eg [5 marks] Examiners report 7.[5 marks] Markscheme METHOD 1 23(r-1) 4 g2X.
  5. Note: Award MI for writing in terms of 2 and 3. 23r X 2 —3 _ 23r X 33r 2-3 = Al In (2—3) = In (33x) (Ml) -31112 = 3r1n3A1 ln3 Al METHOD 2 1118t 1 = In 63r (Ml) (c - 1) ln23 = 3c1n(2 x 3) 3C1n2-31n2=3c1n2+3x1n3Al In 2 iRä Al METHOD 3 1118t 1 In 63r (Ml) (a: — 1)ln8 3xln6A1 In 8 ln8-31n6 Al ln3 Al [5 marks] Examiners report [N/A] 5

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