Exponential and Logarithmic Equations

Lathi, B.P. “Ordinary Linear Differential and Difference Equations” Digital Signal Processing Handbook Ed. Vijay K. Madisetti and Douglas B. Williams Boca Raton: CRC Press LLC, 1999 c  1999byCRCPressLLC 2 Ordinary Linear Differential and Difference Equations B.P. Lathi California State University, Sacramento 2.1 Differential Equations Classical Solution • MethodofConvolution 2.2 Difference Equations Initial Conditions andIterativeSolution • Classical Solution • MethodofConvolution References 2.1 Differential Equations Afunctioncontainingvariablesandtheirderivativesiscalledadifferentialex pression,andanequation involvingdifferentialexpressionsiscalledadifferentialequation. Adifferentialequationisanordinary differential equation if it contains only one independent variable; it is a partial differential equation if it contains more than one independentvariable. Weshall deal here only withordinary differential equations. In the mathematical texts, the independent variable is generally x, which can be anything such as time, distance, velocity, pressure, and so on. In most of the applications in control systems, the independent variable is time. For this reason we shall use here independent variable t for time, although it canstand for any other variable as well. The following equation  d 2 y dt 2  4 + 3 dy dt + 5y 2 (t) = sint is an ordinary differential equation of second order because the highest derivative is of the second order. An nth-order differential equation is linear ifit is of the form a n (t) d n y dt n + a n−1 (t) d n−1 y dt n−1 +···+a 1 (t) dy dt + a 0 (t)y(t) = r(t) (2.1) where the coefficients a i (t) are not functions of y(t). If these coefficients (a i ) are constants, the equation is linear with constant coefficients. Many engineering (as well as nonengineering) systems can be modeled by these equations. Systems modeled by these equations are known as linear time- invariant (LTI) systems. In this chapter we shall deal exclusively with linear differential equations with constant coefficients. Certain other forms of differential equations are dealt with elsewhere in this volume. c  1999 by CRC Press LLC Role of Auxiliary Conditions in Solution of Differential Equations We now show that a differential equation does not, in general, have a unique solution unless some additional constraints (or conditions) on the solution are known. This fact should not come as a surprise. A function y(t) has a unique derivative dy/dt, but for a given derivative dy/dt there are infinite possible functions y(t).Ifwearegivendy/dt , it is impossible to determine y(t) uniquely unless an additional piece of information about y(t) is given. For example, the solution of a differential equation dy dt = 2 (2.2) obtained by integrating both sides of the equation is y(t) = 2t + c (2.3) for any value of c. Equation 2.2 specifies a function whose slope is 2 for all t. Any straight line with a slope of 2 satisfies this equation. Clearly the solution is not unique, but if we place an additional constraint on the solution y(t), then we specify a unique solution. For example, suppose we require that y(0) = 5; then out of all the possible solutions available, only one function has a slope of 2 and an intercept with the vertical axis at 5. By setting t = 0 in Equation 2.3 and substituting y(0) = 5 in the same equation, we obtain y(0) = 5 = c and y(t) = 2t + 5 which is the unique solution satisfying both Equation 2.2 and the constraint y(0) = 5. Inconclusion, differentiation isanirreversibleoperationduringwhichcertain informationislost. Toreversethisoperation,onepieceofinformationabouty(t)mustbeprovidedtorestoretheoriginal y(t). Usingasimilarargument,wecanshowthat,givend 2 y/dt 2 ,wecandeterminey(t)uniquelyonly if two additional Exponential and Logarithmic Equations Exponential and Logarithmic Equations By: OpenStaxCollege Wild rabbits in Australia The rabbit population grew so quickly in Australia that the event became known as the “rabbit plague.” (credit: Richard Taylor, Flickr) In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting Because Australia had few predators and ample food, the rabbit population exploded In fewer than ten years, the rabbit population numbered in the millions Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth In this section, we will learn techniques for solving exponential functions 1/28 Exponential and Logarithmic Equations Using Like Bases to Solve Exponential Equations The first technique involves two functions with like bases Recall that the one-to-one property of exponential functions tells us that, for any real numbers b, S, and T, where b > 0, b ≠ 1, bS = bT if and only if S = T In other words, when an exponential equation has the same base on each side, the exponents must be equal This also applies when the exponents are algebraic expressions Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown 32x For example, consider the equation 34x − = To solve for x, we use the division property of exponents to rewrite the right side so that both sides have the common base, Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x : 32x = 34x − = 34x − = 32x − Use the division property of exponents 4x − = 2x − Apply the one-to-one property of exponents =6 Subtract 2x and add to both sides 4x − 2x 32x 31 Rewrite as 31 x =3 Divide by A General Note Using the One-to-One Property of Exponential Functions to Solve Exponential Equations For any algebraic expressions S and T, and any positive real number b ≠ 1, bS = bT if and only if S = T How To Given an exponential equation with the form bS = bT, where S and T are algebraic expressions with an unknown, solve for the unknown Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT 2/28 Exponential and Logarithmic Equations Use the one-to-one property to set the exponents equal Solve the resulting equation, S = T, for the unknown Solving an Exponential Equation with a Common Base Solve 2x − = 22x − 2x − = 22x − The common base is x − = 2x − By the one-to-one property the exponents must be equal x=3 Solve for x Try It Solve 52x = 53x + x= −2 Rewriting Equations So All Powers Have the Same Base Sometimes the common base for an exponential equation is not explicitly shown In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property For example, consider the equation 256 = 4x − We can rewrite both sides of this equation as a power of Then we apply the rules of exponents, along with the one-toone property, to solve for x : 256 = 4x − 28 = (22) x−5 28 = 22x − 10 = 2x − 10 18 = 2x x=9 How To Rewrite each side as a power with base Use the one-to-one property of exponents Apply the one-to-one property of exponents Add 10 to both sides Divide by Given an exponential equation with unlike bases, use the one-to-one property to solve it Rewrite each side in the equation as a power with a common base Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT 3/28 Exponential and Logarithmic Equations Use the one-to-one property to set the exponents equal Solve the resulting equation, S = T, for the unknown Solving Equations by Rewriting Them to Have a Common Base Solve 8x + = 16x + 8x + = 16x + (23) x+2 = (24) x+1 23x + = 24x + Write and 16 as powers of To take a power of a power, multiply exponents 3x + = 4x + Use the one-to-one property to set the exponents equal x=2 Solve for x Try It Solve 52x = 253x + x= −1 Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base Solve 25x = √2 25x = 2 5x = x= Write the square root of as a power of 2 Use the one-to-one property 10 Solve for x Try It Solve 5x = √5 x= Q&A Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process? 4/28 Exponential and Logarithmic Equations No Recall that the range of an exponential function is always positive While solving the equation, we may obtain an expression that is undefined Solving an Equation with Positive and Negative Powers Solve 3x + = −2 This equation has no solution There is no real value of x that will make the equation a true statement because ...Annals of Mathematics Quasilinear and Hessian equations of Lane-Emden type By Nguyen Cong Phuc and Igor E. Verbitsky* Annals of Mathematics, 168 (2008), 859–914 Quasilinear and Hessian equations of Lane-Emden type By Nguyen Cong Phuc and Igor E. Verbitsky* Abstract The existence problem is solved, and global pointwise estimates of solu- tions are obtained for quasilinear and Hessian equations of Lane-Emden type, including the following two model problems: −∆ p u = u q + µ, F k [−u] = u q + µ, u ≥ 0, on R n , or on a bounded domain Ω ⊂ R n . Here ∆ p is the p-Laplacian defined by ∆ p u = div (∇u|∇u| p−2 ), and F k [u] is the k-Hessian defined as the sum of k × k principal minors of the Hessian matrix D 2 u (k = 1, 2, . . . , n); µ is a nonnegative measurable function (or measure) on Ω. The solvability of these classes of equations in the renormalized (entropy) or viscosity sense has been an open problem even for good data µ ∈ L s (Ω), s > 1. Such results are deduced from our existence criteria with the sharp exponents s = n(q−p+1) pq for the first equation, and s = n(q−k) 2kq for the second one. Furthermore, a complete characterization of removable singularities is given. Our methods are based on systematic use of Wolff’s potentials, dyadic models, and nonlinear trace inequalities. We make use of recent advances in potential theory and PDE due to Kilpel¨ainen and Mal´y, Trudinger and Wang, and Labutin. This enables us to treat singular solutions, nonlocal operators, and distributed singularities, and develop the theory simultaneously for quasi- linear equations and equations of Monge-Amp`ere type. 1. Introduction We study a class of quasilinear and fully nonlinear equations and in- equalities with nonlinear source terms, which appear in such diverse areas as quasi-regular mappings, non-Newtonian fluids, reaction-diffusion problems, and stochastic control. In particular, the following two model equations are of *N. P. was supported in part by NSF Grants DMS-0070623 and DMS-0244515. I. V. was supported in part by NSF Grant DMS-0070623. 860 NGUYEN CONG PHUC AND IGOR E. VERBITSKY substantial interest: (1.1) −∆ p u = f (x, u), F k [−u] = f(x, u), on R n , or on a bounded domain Ω ⊂ R n , where f(x, u) is a nonnegative func- tion, convex and nondecreasing in u for u ≥ 0. Here ∆ p u = div (∇u |∇u| p−2 ) is the p-Laplacian (p > 1), and F k [u] is the k-Hessian (k = 1, 2, . . . , n) defined by (1.2) F k [u] =  1≤i 1 <···<i k ≤n λ i 1 ···λ i k , where λ 1 , . . . , λ n are the eigenvalues of the Hessian matrix D 2 u. In other words, F k [u] is the sum of the k ×k principal minors of D 2 u, which coincides with the Laplacian F 1 [u] = ∆u if k = 1, and the Monge–Amp`ere operator F n [u] = det (D 2 u) if k = n. The form in which we write the second equation in (1.1) is chosen only for the sake of convenience, in order to emphasize the profound analogy be- tween the quasilinear and Hessian equations. Obviously, it may be stated as (−1) k F k [u] = f(x, u), u ≥ 0, or F k [u] = f(x, −u), u ≤ 0. The existence and regularity theory, local and global estimates of sub- and super-solutions, the Wiener criterion, and Harnack inequalities associated with the p-Laplacian, as well as more general quasilinear operators, can be found in Stochastic Mechanics Applications of Random Media Mathematics signal Processing Stochastic Modelling and Image Synthesis and Applied Probability Mathematical Economics and Finance Stochastic Optimization 21 Stochastic Control Stochastic Models in Life Sciences Edited by B. Rozovskii M. Yor Advisory Board D. Dawson D. Geman G. Grimmett I. Karatzas F. Kelly Y. Le Jan B. Bksendal E. Pardoux G. 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Hamiltonian Systems and HJB Equations (1999) 44 MỤC LỤC Mục lục Chuyên đề Công thức mũ, lũy thừa logarit 1.1 Công thức mũ lũy thừa 1.2 Công thức logarit Chuyên đề Hàm số lũy thừa, hàm số mũ hàm số logarit 11 2.1 Tập xác định hàm số 11 2.2 Đạo hàm giá trị nhỏ nhất, giá trị lớn hàm số - Tiếp tuyến đồ thị hàm số 13 2.3 Tính đơn điệu, cực trị đồ thị hàm số 15 Chuyên đề Phương trình mũ phương trình logarit 23 3.1 Phương trình mũ logarit 23 3.2 Phương pháp đưa số 24 3.3 Phương pháp logarit hóa mũ hóa 25 3.4 Phương pháp đặt ẩn phụ 26 CHUYÊN ĐỀ CÁC CÔNG THỨC MŨ - LŨY THỪA VÀ LOGARIT 1.1 Công thức mũ lũy thừa 1.1.1 Bài tập tự luận 1.1.2 Câu hỏi trắc nghiệm khách quan Cấp độ nhận biết thông hiểu Câu Tính giá trị biểu thức A = 625 A 14 B 12 Câu Kết phép tính A = 16 A 40 B 32 −1 + 16 − 2−2 64 C 11 −0,75 D 10 + 0, 25− C −24 −0,25 Câu Kết phép tính B = 27 + − 250,5 16 A B C 16 D 257 D 54 Câu Cho x, y hai số thực dương m, n hai số thực tùy ý Đẳng thức sau sai? n m A (x ) = x n.m m n B x x = x m+n xm C n = y Câu Cho a, b > 0; m, n ∈ N∗ Hãy tìm khẳng định đúng? x y m−n D (xy)n = xn y n Th.S Trần Quang Thạnh √ n Sđt: 0935-29-55-30 m B an : bm = (a : b)m−n am = a n √ √ n k a = n+k a C A D an bn = (a.b)n √ √ Câu Rút gọn biểu thức P = a a √ 3+1 A P = a B P = a 3−1 3+2 với a > √ C P = a2 3+1 D P = a 2√ Câu Cho a số thực dương, biểu thức a a viết dạng lũy thừa với số mũ hữu tỉ A a B a Câu Cho f (x) = D a √ √ x x Khi f (0, 09) A 0,1 B 0,2 Câu Biểu thức 11 C a C 0,3 D 0,4 √ √ √ x x x5 , x > viết dạng lũy thừa với số mũ hữu tỉ A x B x C x D x √ 4 a3 b2 Câu 10 Rút gọn √ , với a,b số thực dương ta a12 b6 A a2 b B ab2 C a2 b2 Câu 11 Cho biểu thức A = (a + 1)−1 + (b + 1)−1 Nếu a = + D a.b √ −1 b = − √ −1 giá trị A A B C D √ Câu 12 Cho biểu thức P = x x x x, x > Mệnh đề đúng? A P = x Câu 13 Biểu thức C = 15 B P = x 10 15 B x Câu 14 Cho biểu thức D = A D = x 13 D x 16 √ x x2 x3 , với x > Mệnh đề đúng? B D = x 24 Câu 16 Cho biểu thức P = C x 16 √ a Câu 15 Rút gọn biểu thức E = (1 + a2 )−1 √ √ A B 2a A P = x D P = x √ x x x x (x > 0) viết dạng lũy thừa số mũ hữu tỉ A x 18 13 C P = x 10 C D = x D D = x √ 2 − a−2 − −1 : (với a = 0, a = ±1) a a−3 C a D a √ x x2 x3 , với x > Mệnh đề đúng? 13 B P = x 24 C P = x D P = x Câu 17 Cho số thực a, b, α (a > b > 0, α = 1) Mệnh đề sau đúng? a α aα A (a + b)α = aα + bα B = −α C (a − b)α = aα − bα D (ab)α = aα bα b b AMS-LATEX Trang Th.S Trần Quang Thạnh Sđt: 0935-29-55-30 √ 4 a3 b2 Câu 18 Cho a, b số dương Rút gọn biểu thức P = √ kết a12 b6 A ab2 B a2 b C ab D a2 b2 1√ 1√ a3 b + b3 a √ √ Câu 19 Cho số thực dương a b Rút gọn biểu thức P = √ − ab 6 a+ b A B −1 C D −2 Câu 20 Cho số thực dương a Biểu thức thu gọn biểu thức P = A B a + C 2a −n −n a a− + a 3 là: a a + a− D a −n a +b a − b−n − (với ab = 0, a = ±b) −n − b−n −n + b−n a a a n bn 2an bn 3an bn 4an bn A 2n B C D b − a2n b2n − a2n b2n − a2n b2n − a2n √ Câu 22 Cho a > Viết biểu thức P = a a6 dạng lũy thừa với số mũ hữu tỷ Câu 21 Rút gọn biểu thức F = A P = C P = a7 B P = a D P = a6 Câu 23 Trong khẳng định sau, khẳng định sai? A Nếu a > ax > ay x > y B Nếu a > ax ≤ ay x ≤ y C Nếu < a < ax > ay x > y D Nếu < a = ax = ay x = y 7 x y + x.y Câu 24 Cho x, y > 0, rút gọn P = √ √ x+ 6y √ √ C P = x.y A P = x + y B P = x + y √ Câu 25 Cho a > 0, rút gọn P = a √ 5−2 √ a1− a B P = a √ D P = √ xy 5+2 3−2 C P = D P = a2 a √ Câu 26 Giả sử a số thực dương, khác Biểu thức a a viết dạng aα Khi 11 A α = B α = C α = D α = 3 √ Câu 27 Đưa biểu thức A = a a a lũy thừa số < a = 1ta biểu thức A P = ... instruction and practice with exponential and logarithmic equations • Solving Logarithmic Equations • Solving Exponential Equations with Logarithms 15/28 Exponential and Logarithmic Equations Key Equations. .. Exponential and Logarithmic Equations Solving Applied Problems Using Exponential and Logarithmic Equations In previous sections, we learned the properties and rules for both exponential and logarithmic. .. 11/28 Exponential and Logarithmic Equations Using the One-to-One Property of Logarithms to Solve Logarithmic Equations As with exponential equations, we can use the one-to-one property to solve logarithmic