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184 Section 6.6 – Logarithmic and Exponential Equations In this section, we will use the following tools to help us solve exponential and logarithmic equations: Tool 1: Loga Definition For x > 0 and a > 0 and a ≠ 1, then y = f(x) = loga(x) if and only if x = ay. Tool 2: Loga is 1-to-1 Theorem Let M, N, and a be positive real numbers and a ≠ 1, then M = N if and only if loga(M) = loga(N). Tool 3: ax is 1-to-1 Theorem Let u and v be real numbers and a > 0 and a ≠ 1, then u = v if and only if au = av We will also use our properties of logarithms to help solve equations: Tool 4: Law of Logs Theorem Let M > 0, N > 0, a > 0, a ≠ 1, and r be any real number. Then a) loga(M•N) = loga(M) + loga(N) ( MN ) = log (M) – log (N) b) loga c) loga(Mr) = r loga(M) a a We will need to be cautious in using these properties since that even though € log of a product and/or quotient may be defined, the log of each factor may not be, so we will need to check our answers with the original equation. Also, using loga(Mr) = r loga(M) when r is an even integer can leave out possible solutions or introduce extraneous ones. Objective #1: Solve Logarithmic Equations. To solve logarithmic equations, we need to get a our equation into either the form of loga(M) = loga(N) and apply the Loga is 1-to-1 theorem or the form y = loga(x) and apply the Loga definition. Solve the following: Ex. 1 log5(3 – 4x) = 2 185 Solution: Since log5(3 – 4x) is defined when 3 – 4x > 0 or x < ¾, then the domain is (– ∞, ¾). Using the Loga Definition, we can rewrite the equation as an exponential equation: log5(3 – 4x) = 2 (apply the Loga Definition) 2 3 – 4x = 5 3 – 4x = 25 (solve for x) – 4x = 22 x = – 5.5 Since – 5.5 is in the domain of the function, the solution is {– 5.5}. Ex. 2 – 2 log6(x) = log6(16) Solution: Since log6(x) is defined when x > 0, then the domain is (0, ∞). We can rewrite – 2 log6(x) as log6(x – 2) using The Law of Logs, part c. – 2 log6(x) = log6(16) (apply The Law of Logs, part c) log6(x – 2) = log6(16) (apply the Loga is 1-to-1 Thm) –2 x = 16 (take the "reciprocal of both sides") 1 2 x = (use the square root property) x= 16 1 ± 4 1 Since the domain is (0, ∞), our only solution is { }. 4 € Ex. 3 log49(6x + 1) + log49(x + 2) = ½ € Solution: Since log49(6x + 1) is defined when 6x € + 1 > 0 or x > – 1 6 and since log49(x + 2) is defined when x + 2 > 0 or x > – 2, the domain is 1 1 (– , ∞) ∩ (– 2, ∞) = (– , ∞). 6 € 6 log49(6x + 1) + log49(x + 2) = ½ (apply € the Law of Logs, part a) log49[(6x + 1)(x + 2)] = ½ (FOIL) 2 log49[6x + 13x + 2] = ½ (apply the Log Definition) 2 1/2 € 6x + 13x + 2 = 49 (simplify) 2 6x + 13x + 2 = 7 (subtract 7 from both sides) 2 6x + 13x – 5 = 0 (factor) (2x + 5)(3x – 1) = 0 (solve) 5 1 x=– or x = 2 Since the domain is (– € 3 1 , 6 1 3 ∞), our only solution is { }. € € € 186 Ex. 4 ln(3x) – ln(x – 8) = ln(x + 5) Solution: Since ln(3x) is defined when x > 0, ln(x – 8) is defined when x – 8 > 0 or when x > 8, and ln(x + 5) is defined when x + 5 > 0 or x > – 5, the domain is (0, ∞) ∩ (8, ∞) ∩ (– 5, ∞) = (8, ∞). ln(3x) – ln(x – 8) = ln(x + 5) (apply the Law of Logs, part b) ln 3x ( x−8 ) = ln(x + 5) 3x x−8 =x+5 (apply the Loga is 1-to-1 Thm) (multiply both sides by x – 8) 3x = (x + 5)(x – 8) (FOIL) 2 € 3x = x – 3x – 40 (subtract 3x from both sides) 2 0 = x – 6x – 40 (factor) € 0 = (x – 10)(x + 4) (solve) x = 10 or x = – 4 But, x = – 4 is not in the domain, so the solution is {10}. Ex. 5 log4(x2) = 2 Solution: Since the log4(x2) is defined when x2 > 0 or when x ≠ 0, the domain is (– ∞, 0) U (0, ∞). log4(x2) = 2 (apply the Log Definition) 2 2 x =4 (simplify) 2 x = 16 (use the square root property) x=±4 Both values are in the domain, so the solution is {– 4, 4}. Suppose we had use the Law of Logs, part c as our first step. We would have arrived at the following solution: log4(x2) = 2 2 log4(x) = 2 (divide both sides by 2) log4(x) = 1 (apply the Log Definition) 1 x = 4 = 4 (missing an answer) Hence, notice that does not gives us all of the answers. That is why it is important to determine the domain before you begin and use great caution when using the Law of Logs, part c. Objective #2: Solve Exponential Equations. To solve exponential equations, we try to get our equations in either the form au = av and apply the ax is 1-to-1 theorem or the form au = M and apply the loga is 1-to-1 by taking the natural log of both sides. 187 Solve the following. Give both the exact and approximate answers when appropriate: Ex. 6 3•2x – 5 = 48 Solution: The domain is all real numbers, so we have no restrictions. First, divide both sides by 3 to get 2x – 5 = 16. Since 16 = 24, rewrite the equation as: 2x – 5 = 24 (apply the ax is 1-to-1 thm.) x–5=4 (solve) x=9 The solution is {9}. Ex. 7 4 – x – 2.8 = 0 Solution: The domain is all real numbers, so we have no restrictions. 4 – x – 2.8 = 0 (add 2.8 to both sides) –x 4 = 2.8 (use Loga is 1-to-1 thm by taking ln of both sides) –x ln(4 ) = ln(2.8) (apply the Law of Logs, part c) – x ln(4) = ln(2.8) (solve for x) ln(2.8) x=– ≈ – 0.7427 ln(4) { The solution is – Ex. 8 ln(2.8) ln(4) } ≈ {– 0.7427} € 23x – 4 = 71 + 2x Solution: € is all real numbers, so we have no restrictions. The domain 23x – 4 = 71 + 2x (use Loga is 1-to-1 thm by taking ln of both sides) 3x – 4 1 + 2x ln(2 ) = ln(7 ) (apply the Law of Logs, part c) (3x – 4) ln(2) = (1 + 2x) ln(7) (distribute) 3x ln(2) – 4 ln(2) = ln(7) + 2x ln(7) (add 4 ln(2) – 2x ln(7) to both sides) 3x ln(2) – 2x ln(7) = ln(7) + 4 ln(2) (factor x out) [3 ln(2) – 2 ln(7)]x = ln(7) + 4 ln(2) (divide both sides by 3 ln(2) – 2 ln(7)) ln(7)+4ln(2) x= ≈ – 2.6035 The 3ln(2)−2ln(7) ln(7)+4ln(2) solution is 3ln(2)−2ln(7) { € € } ≈ {– 2.6035}. 188 Ex. 9 32x – 3x + 2 – 10 = 0 Solution: The domain is all real numbers, so we have no restrictions. This equation almost appears to be quadratic since 32x = (3x)2 and 3x + 2 = 3x•32 or 9•3x. Let's try letting w = 3x. Thus, we get: 32x – 3x + 2 – 10 = 0 (3x)2 – 9•3x – 10 = 0 (replace 3x by w) w2 – 9w – 10 = 0 (factor) (w – 10)(w + 1) = 0 (solve for w) w = 10 or w = – 1 (replace w by 3x) 3x = 10 or 3x = – 1 (take the ln of both sides) x x ln(3 ) = ln(10) or ln(3 ) = ln(– 1) Since ln(– 1) is undefined, the second equation has no solution. ln(3x) = ln(10) (apply the Laws of Logs, part c) x ln(3) = ln(10) (solve for x) ln(10) x= ≈ 2.0959 ln(3) ≈ {2.0959}. { ln(10) ln(3) } The solution is log2(x) = 2x Ex. 10a € Ex. 10b log1/2(x) = 1 x 2 ( ) Solution: € This problem is extremely difficult to solve algebraically, but very easy if we remember the graphs of y = loga(x) and y = ax: € 4 4 3 2 y=2 y= x 1 x ( 21 ) 2 1 y = log2(x) 0 -4 -3 -2 -1 -1 0 0 1 2 3 4 € -4 -3 -2 -1 -1 -2 -2 -3 -3 -4 y = log1/2(x) 3 x 0 1 2 -4 Thus, log2(x) = 2 has no solution and the solution to log1/2(x) = is ≈ {2/3}. € 3 1 x 2 ( ) 4