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Elementary Algebra

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elroy
16 diapositives · 471 Views · Sep 10, 2014

Elementary Algebra. Section 4.3 Properties of Logarithms. Properties of Logarithms. Consider log b x = m and log b y = n By definition b m = x and b n = y xy = (b m )(b n ) = b m + n So log b (xy) = log b (b m + n ) Product Rule for Logarithms

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Elementary Algebra
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  1. Elementary Algebra Section 4.3 Properties of Logarithms

  2. Properties of Logarithms • Consider logb x = m and logb y = n By definition bm = x and bn = y xy = (bm)(bn) = bm + n So logb (xy) = logb (bm + n) Product Rule for Logarithms logb xy = logb x + logb y for any positive real numbers b, x, y with b ≠ 1 = m + n WHY? = logb x + logb y Section 4.5 v5.0.1

  3. ( ) = ln (x – 1)(x + 1) Properties of Logarithms • Examples • 1. log4 (3  7) = log4 3 + log4 7 • 2. log8 10 + log8 3 = log8 (10  3) = log8 30 • 3. loga x2 = loga xx = loga x + loga x = 2 loga x • 4. 1 + log 2 + log x + log 2x2 • 5. ln (x – 1) + ln (x + 1) = 1 + log 4x3 = ln (x2 – 1) Section 4.5 v5.0.1

  4. x y bm logb bm – n m – n logb x – logb y = = = logb logb = bn x logb = logb x – logb y y Properties of Logarithms • Again consider logb x = m and logb y = n , for x, y, b positive, b ≠ 1 Thus bm = x and bn = y Quotient Rule for Logarithms for any positive real numbers b, x, y with b ≠ 1 WHY? Section 4.5 v5.0.1

  5. log (x + 3) 2 = log (x + 1) log (x + 1) log (x + 1) = 2 2 = (x + 1) x2+ 2x+ 1 x + 3 = = 2 ( ) (x + 2)(x – 1) = log7 9 Solution set: x2+ x – 2 0 = { –2 , 1 } 4 ( ) 3 log4 16 Properties of Logarithms • Examples • 1. • 2. • 3. Solve for x: = log7 9 – log7 4 = log4 3 – log4 16 = log4 3 – 2 WHY? log (x + 3) WHY? Section 4.5 v5.0.1

  6. Properties of Logarithms • Consider logb x = m for x, b positive, b ≠ 1 Thus bm = x and xr = bmr So logb xr=logb (brm) Power Rule for Logarithms for any positive real numbers b, x with b ≠ 1 and (bm)r = xr , for any realr = brm = rm WHY? = r logb x logb xr = r logb x Section 4.5 v5.0.1

  7. Properties of Logarithms • Examples • 1. • 2. • 3. 3 log5(x + 1) • 4.x log 2 • 5.ln 1 • 6. ln 0 log352 = 2 log35 = 4 loga x loga x4 = log5(x + 1)3 = log 2x = loge1 = 0 = ? Question: Is 0 in the domain of any logarithm function ? What does this tell you about ln 0 ? Section 4.5 v5.0.1

  8. ( ) 1 20e ( ) e 20 logb x logb y Properties of Logarithms • Property Recognition Rewrite as a logarithm of a single expression : 1. log 4 + log 7 2. log 35 – log 7 3. ln 5e – ln 4. log 5e – log 5. logb x5 – logb x3+ logb x2 6. 7. logb (x  y) in terms of logb x and logb y Section 4.5 v5.0.1

  9. Properties of Logarithms • More Examples 1. Rewrite in expanded form: log4 (3x + 7) Cannot be written in expanded form ! 2. TRUE or FALSE : log6 36 – log6 6 = log6 30 Rewriting: log6 (36/6) = log6 (6  5) log6 6 = log6 6 + log6 5 0 = log6 5 Since log6 1 = 0 then log6 1 = log6 5 This implies that 1 = 5 ... a CONTRADICTION !! Hence the given statement is FALSE ! Section 4.5 v5.0.1

  10. log3 (log2 8) = log3 (3) = 2 log7 7 log7 49 log7 72 1 = 2 log8 8 log8 64 log8 82 2 (1) 1 = 2 (1) 1 = 1 Properties of Logarithms • More Examples 3. TRUE or FALSE : So, the given statement is TRUE !! Section 4.5 v5.0.1

  11. replacing y applyinglogb ... applying power rule loga x logb x = = logb x loga x logb x logba loga b logba logb x – logba ≠ Bases for Logarithms • Conversions Can we use logb x to find loga x ? Let loga x = y By definition ay = x logb (ay) = logb x y logba = logb x (loga x)(logba) = logb x Thus OR NOTE: Section 4.5 v5.0.1

  12. 1.23044 2.5789 2.5789 ≈ ≈ = = 0.477121 2.8332 = 1.0986 ln 17 = ln 3 log 17 log 3 Bases for Logarithms • Conversion Examples Find log3 17 on your calculator Having trouble ? ... if you can Let’s try using a little math first ... log3 17 OR log3 17 Section 4.5 v5.0.1

  13. { 263 } Solution set: = 33 3log (x+1) 5 3 { –1 + 271/5} Solution set: More Equations • Solve 1. Find x to the nearest whole number e.02x = 192 ln(e.02x) = ln(192) (.02x)ln e = ln(192) .02x = 5.2575 x≈ 262.9 ≈ 263 2. Find xexactly log3 (x + 1)5 = 3 (x + 1)5 = 27 x + 1 = 271/5 x = –1 + 271/5 Section 4.5 v5.0.1

  14. { 3 } Solution set: ( ) 2x 2x –2 log3 = 3–2 = 3x + 15 3x + 15 { 1 } Solution set: More Equations • Solve 3. Find x exactly log8(2x + 5) + log8 3 = log8 33 log8(2x + 5) + log8 3 = log8 3 + log8 11 log8(2x + 5) = log8 11 2x + 5 = 11 x= 3 4. Find xexactly log3 2x – log3 (3x + 15) = –2 = log8 (3 ∙ 11) 18x = 3x + 15 x = 1 Section 4.5 v5.0.1

  15. Think about it ! Section 4.5 v5.0.1

  16. Module 4 Section 4.5 Properties of Logarithms