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Today in Precalculus

Learn the step-by-step process for solving log equations in precalculus. Understand how to isolate log terms, apply properties of logs, convert to exponents, and check for extraneous solutions. Practice examples provided.

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Today in Precalculus

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  1. Today in Precalculus • Go over homework • Notes: Solving Log Equations • Homework • Quiz Friday, January 8

  2. Steps to Solving Log Equations • If there is only one term with a log or LN term, isolate it. • If there is more than one term on either side with a log term, use properties of logs to condense it to one term. • Convert to an exponent (if there is ONE log term on each side, can cancel the log) • Use algebra to solve for unknown. • Check for extraneous solutions by substituting solutions into original equation and verify no negatives values in log term.

  3. Example lnx = 8 e8 = x x = 2980.96 Check: ln2980.96 =8 so x = 2980.96 is the answer

  4. Example Check: 7 + 3log62(1.65) 7 + 3log63.30 which is positive, so answer is x = 1.65 7 + 3log62x = 9 3log62x = 2 x = 1.65

  5. Example log(4x + 2) – log(x – 1) = 1 10x – 10 = 4x + 2 x = 2 Check: both log terms still positive, so x=2 is the answer

  6. Example ln(x – 2) + ln(2x – 3) = 2lnx ln(x – 2)(2x – 3) = lnx2 ln(2x2 – 7x + 6) = lnx2 2x2 – 7x + 6 = x2 x2 – 7x + 6 = 0 (x – 6)(x – 1) = 0 x – 6 = 0 x – 1 = 0 x=6 x = 1 Check: ln(1 – 2) is ln(-1) so 1 is an extraneous solution, only answer is x = 6

  7. Example logx2 = 2 102 = x2 100 = x2 x = -10, 10

  8. Homework • Pg 331: 7-10, 17,18,25-28,35-38 • Quiz: Friday, January 8

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