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Solving Exponential Equations

Solving Exponential Equations. Example 1 10 x-2 = (2) -x+4 log10 x-2 = log2 -x+4 (x-2)log10 = (-x+4)log2 (x-2) log10 = (-x+4) log2 3.32(x-2) = (-x+4) 3.32x-6.64 = -x+4 4.32x = 10.64 x = 2.46. Example 2 2 x+1 = 3 5 x

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Solving Exponential Equations

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  1. Solving Exponential Equations Example 1 10x-2 = (2)-x+4 log10x-2 = log2-x+4 (x-2)log10= (-x+4)log2 (x-2)log10= (-x+4) log2 3.32(x-2)= (-x+4) 3.32x-6.64= -x+4 4.32x = 10.64 x = 2.46

  2. Example 2 2x+1= 3 5x log 2x+1= log3 5x log 2x+1 -log 5x = log3 (x+1)log 2-xlog 5= log3 xlog 2 + log 2-xlog 5= log3 xlog 2 -xlog 5= log3 - log 2 xlog2 = log3 5 2 x = -0.4425

  3. Solving Log Equations*Find the RESTRICTIONS of the argument* Example 3 2log5x = 4 log5x = 2 102 = 5x Restriction: 5x > 0 so x > 0 100 = x 5 x = 20

  4. Example 4 log(x+1) = 1 – log(x-2) log(x+1) + log(x-2) = 1 log((x+1)(x-2)) = 1 101 = (x+1)(x-2) Restrictions:x+1 > 0 x–2 > 0 x > -1 and x > 2

  5. x2 – x – 2 = 10 x2 – x – 12 = 0 (x+3)(x-4) = 0 x = -3 and x = 4 Because of the restrictions, we reject x = -3 S={4}

  6. Example 5 ln(x+2) = ln(-x+6) + ln(x-1) Restrictions: x > -2, x < 6, x >1 ln(x+2) = ln((-x+6)(x-1)) (x+2) = (-x+6)(x-1)(divide both sides by ln) x+2 = -x2 + 7x - 6(expand) -x >-6 x<6

  7. x2 – 6x + 8 = 0 (x-2)(x-4) = 0(factoring) x=2 and x=4 Both fit the restrictions S={2,4}

  8. Solving Log Inequalities • Determine the restriction(s) (make argument > 0) • Solveinequality as an equation • Place values on a number line and use test points

  9. -1 -0.5 1 -0.75 0 Example 6 -18log9(-4x) ≥ -9 Restrictions: -4x > 0 so x < 0 -18log9(-4x) = -9 log9(-4x) = 0.5 90.5 = (-4x)3 = -4x x = -0.75 If x = -1 -11.4 ≥ -9 If x = -0.5 -5.7 ≥-9 If x = 1 No need to check x ε[ -0.75, 0 [

  10. 0 10 36 8 35 Example 7 log9(x – 8) > 1.5 Restrictions: x - 8 > 0 so x > 8 log9(x-8) = 1.5 91.5 = (x-8)27 = x-8 x = 35 If x = 0 No need to check If x = 10 0.32 >1.5 If x = 36 1.53 > 1.5 x ε] 35, +∞[

  11. 0 0.75 2 2.75 4 0.5 1 2.5 3 Example 8 ln(x – 1) + ln(2x-5) ≤ ln2 Restrictions: x > 1 and x > 2.5 ln(x – 1) + ln(2x-5) = ln2 ln[(x – 1)(2x-5)] = ln2 2x2 -7x + 5 = 2 2x2 -7x + 3 = 0 x = 0.5 and x = 3 If x = 0 No need to check If x = 0.75 No need to check If x = 2 No need to check If x = 2.75 -0.13 ≤ 0.69 If x = 4 2.19 ≤ 0.69 x ε]2.5, 3]

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