5.6 Laws of Logarithms 5.7 Exponential Equations; Changing Base Objectives: 1.Compare & recall...

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Transcript of 5.6 Laws of Logarithms 5.7 Exponential Equations; Changing Base Objectives: 1.Compare & recall...

  • 5.6 Laws of Logarithms5.7 Exponential Equations; Changing BaseObjectives:1.Compare & recall the properties of exponents2.Deduce the properties of logarithms from/by comparing the properties of exponentsUse the properties of logarithmsSolve the exponential equations

  • Since the logarithmic function y = logb x is the inverse of the exponential function y = bx, the laws of logarithms are very closely related to the laws of exponent.Pre-KnowledgeFor any b, c, u, v +, and b 1, c 1, there exists some x, y , such thatu = bx, v = byBy the previous section knowledge, as long as takingx = logbu, y = logbv

  • 1. Product of Power am an = am+n1. Product Propertylogbuv = logbu + logbvProoflogbuv = logb(bxby)= logbb x+y = x + y = logbu + logbv Pre-KnowledgeFor any b, c, u, v +, and b 1, c 1, there exists some x, y , such thatu = bx, v = byBy the previous section knowledge, as long as takingx = logbu, y = logbv

  • 2. Quotient Property2. Quotient of PowerProofPre-KnowledgeFor any b, c, u, v +, and b 1, c 1, there exists some x, y , such thatu = bx, v = byBy the previous section knowledge, as long as takingx = logbu, y = logbv

  • Equal Power am = an iffm = nEqual Property logbu = logbviff u = vProoflogbu logbv = 0

  • 4. Power of Power(am)n = amn 4. Power Propertylogbuk = k logbuProoflogbuk = logb(bx)k = logbb kx = kx = k logbu Pre-KnowledgeFor any b, c, u, v +, and b 1, c 1, there exists some x, y , such thatu = bx, v = byBy the previous section knowledge, as long as takingx = logbu, y = logbv

  • 5. Change-of-Base FormulaProof Note thatbx = u, logbu = xTaking the logarithm with base c at both sides:logcbx = logcu orx logcb = logcu

  • 6. Reciprocal FormulaProof

  • 7. Raise Power FormulaProof

  • Example 1 Assume that log95 = a, log911 = b, evaluatelog9 (5/11)

    log955

    log9125

    log9(121/45)

    log9275

  • Example 2 Expanding the expression

    ln(3y4/x3)ln(3y4/x3) = ln(3y4) lnx3 = ln3 + lny4 lnx3= ln3 + 4 ln|y| 3 lnx

    b) log3125/6x9 log3125/6x9 = log3125/6 + log3x9 = 5/6 log312 + 9 log3x= 5/6 log3(3 22) + 9 log3x= 5/6 (log33 + log322) + 9 log3x= 5/6 ( 1 + 2 log32) + 9 log3x

  • Example 3 Condensing the expressiona) 3 ( ln3 lnx ) + ( lnx ln9 ) 3 ( ln3 lnx ) + ( lnx ln9 ) = 3 ln3 3 lnx + lnx 2 ln3 = ln3 2 lnx = ln(3/x2)

    b) 2 log37 5 log3 x + 6 log9 y2 2 log37 5 log3 x + 6 log9 y2 = log349 log3 x5 + 6 ( log3 y2/ log39)= log3(49/x5) + 3 log3 y2= log3(49y6/x5)

  • Practice

    A) P. 199 Q 7 18 P. 199 Q 19 20 How do you change to make it to be true?

    True or False log a y = log 1/a y

    c) P. 200 Q 7 27 (odd)

  • Example 4 Calculate log48 and log615 using common and natural logarithms. a) log48 log48 = log8 / log4 = 3 log2 / (2 log2)= 3/2 log48 = ln8 / ln4 = 3 ln2 / (2 ln2) = 3/2

    b) log615 = log15 / log6 = 1.511

  • Example 5. Express in terms of

  • More on Expand/Condense logarithmic expressionsExample 6 Expand

  • Example 7 Expand in terms of sums and differences of logarithmsMore on Expand/Condense logarithmic expressions

  • Example 8 Expand to express all powers as factorsMore on Expand/Condense logarithmic expressions

  • Example 9 Condense to a single logarithm. More on Expand/Condense logarithmic expressions

  • Assignment:5.6 P. 196 #36 44 (even)P. 200 #2 22 (even), 21 33 (odd), 41, 43, 45

  • Solving Exponential EquationsOne way to solve exponential equations is to use the property that if 2 powers with the same base are equal, then their exponents are equal.

    For b > 0 and b1if bx = by, then x = y

  • Solve by Equating Exponents Example 10: Solve 43x = 8x+1

    (22)3x = (23)x+1rewrite with same base26x = 23x+36x = 3x + 3x = 1Check 43*1 = 81+1 64 = 64

  • Your turn!Solve: 16x = 32x124x = 32x124x = (25)x14x = 5x 5x = 5Be sure to check your answer!!!

  • When you cant rewrite using the same base, you can solve by taking a log of both sides or use the definition of logExample 11: Solve 2x = 7log22x = log27x = log27

    x = 2.807

  • Example 12: Solve 102x 3 + 4 = 21102x 3 = 17log10102x 3 = log10172x 3 = log 172x = 3 + log17x = (3 + log17) 2.115When you cant rewrite using the same base, you can solve by taking a log of both sides or use the definition of log

  • Solve: 5x + 2 + 3 = 255x+2 = 22log55x+2 = log522x + 2 = log522x = (log522) 2 = (log22/log5) 2 0.079Your turn!

  • Example 13: Solve ex 3e-x = 2More on Solving Exponential Equations[Answer] Multiply ex at both sides of the equation: ex(ex 3e-x ) = 2ex

    e2x 3 = 2ex e2x 2ex 3 = 0(ex)2 2(ex) 3 = 0Denote ex = u, then(u)2 2(u) 3 = 0(u)2 2(u) 3 = 0(u 3)(u + 1) = 0u = 3, or u = 1 ex = 3, or ex = 1 (discard)x = ln3

  • 1. To solve use the property for logs with the same base:b, x, y+ and b 1If logb x = logb y, then x = y

    2. When you cant rewrite both sides as logs with the same base exponentiate each sideb, x+ and b 1if logb x = y, then x = by This can get the expression in the log out of the log simply.Solving Logarithmic Equations

  • Example 14: Newtons Law of Cooling The temperature T of a cooling substance at time t (in minutes) is:T = (T0 TR) e-rt + TRT0= initial temperatureTR= room temperaturer = constant cooling rate of the substance Solving Logarithmic Equations

  • Example 14: Youre cooking stew. When you take it off the stove the temp. is 212F. The room temp. is 70F and the cooling rate of the stew is r = 0.046. How long will it take to cool the stew to a serving temp. of 100?Solving Logarithmic EquationsT = (T0 TR) e -rt + TRT0 = 212, TR = 70, T = 100 r = 0.046So solve:100 = (212 70)e -0.046t + 70

  • 30 = 142e -0.046t (subtract 70)15/71 = e -0.046t(divide by 142)How do you get the variable out of the exponent?Solving Logarithmic Equationsln(15/71) = lne-.046t (take the ln of both sides)ln(15/71) = 0.046tln(15/71)/( 0.046) = tt = (ln15 ln71)/( 0.046) =t 1.556/( 0.046)t 33.8 about 34 minutes to cool!

  • Example 15: Solve log3(5x 1) = log3(x + 7)5x 1 = x + 74x = 8x = 2 and checklog3(52 1) = log3(2 + 7)log39 = log39

    Because the domain of log functions doesnt include all reals, you should check for extraneous solutions.Solving Logarithmic Equations

  • Example 16: Solve log5(3x + 1) = 23x + 1 = 5-23x + 1 = 1/25x = 8/25 and check

    Because the domain of log functions doesnt include all reals, you should check for extraneous solutions.Solving Logarithmic Equations

  • Example 16: Solve log5x + log(x + 1) = 2log[5x(x + 1)]= 2 (product property)log (5x2 + 5x) = 25x2 + 5x = 100x2 + x 20 = 0 (subtract 100 and divide by 5)(x + 5)(x 4) = 0 x = 5, or x = 4

    check and youll see x = 4 is the only solution. More on Solving Logarithmic Equations

  • Your Turn! Solve log2x + log2(x 7) = 3log2 [x(x 7)]= 3log2 (x2 7x) = 3x2 7x = 23x2 7x = 8x2 7x 8 = 0(x 8)(x + 1) = 0x = 8 x = 1 Checklog28 + log2(8 7) =33 + 0 = 3

  • One More!Solve log6(x 2) + log6(x + 3) = 2 log6(x 2) + log6(x + 3) = 2log6 [(x 2)(x + 3)] = 2 log6 (x2 + x 6) = 2x2 + x 6 = 36x2 + x 42 = 0(x 6)(x + 7)=0x = 6 x = 7 Checklog64 + log69 =2log636 = 2Checklog64 + log69 =2log636 = 2

  • Challenge!Example 17: Solve log2x + log4(x2 4x + 4) = 3log4 x2 + log4(x 2)2 = 3 (Raise Power Formula)log4 [x2(x 2)2] = 3x2(x 2)2 = 43 [x(x 2)]2 = 64 A2B2 = (AB)2x(x 2) = 8 A2 = 64, A = 8x2 2x 8 = 0 or x2 2x + 8 =0x = 4 x = 2 No real solution

  • Challenge!Solve log2x + log2(x 1) = 1 2log2x + log2(x 1) = 2log2x2 + log2(x - 1) = 2log2 [x2(x 1)] = 2 x2(x 1) = 4x3 x2 4 = 0(Rational Zero Theorem)(x 2)(x2 + x + 2) = 0x 2 = 0 or x2 + x + 2 = 0x = 2 No real solutionChecklog22 + log21 =12 + 0 = 2

  • Challenge Simplify (No calculator)

    1)

    2)

    3)

    4)

    5) Proof

  • Assignment:5.7 P. 201 #24 34 (even), 42, 44P. 179 #49 52