Question and Answers #37 Fractions

Questions and Answers Sheet Fractions Question #1 Find the Solution of the Exponential Equation? How do I solve 5 = 4 + 1 ? x x I understand how to solve for x when there is one exponent, but I don't know how to solve when there is an exponent on both sides of the equation.. Answer: If a 1 𝑥 𝑎 𝑥 = 𝑒 ln 𝑎 = 𝑒 𝑥 ln 𝑎 ⇒ Therefore a Now, Since x 𝑑𝑎 𝑥 = ln 𝑎 ⋅ 𝑎 𝑥 > 0 𝑑𝑥 and it follows it is injective. 5 5  1( ) x 4 4 , and consequently so is 5 ( )x −1 4 Multiply by the strictly increasing 4 x to get an injective function: 5 [( ) x − 1]  4 x = 5 x − 4 x 4 The conclusion is that there is at most one real x5 − 4 = 1 , which is equivalent to: x x 5x = 4 x + 1 . This single x1 as Tom Bombadil pointed out. Question #2 𝑎 Adding 1 to ? 𝑏 a by finding the least common multiple of the divisors. So I got b 1b a b a ab a + + . So then I added those two fractions together and got I saw that the two b which is 1b b b b b 1 2 a 2 a1 get a , this doesn't seem to be right because 1 + isn't 21 . Am I doing something wrong or is 3 b 3 I've been learning algebra 21 this actually correct, if so can you explain to me why? Answer: 𝑏 𝑎 𝑎𝑏 𝑏 𝑏 𝑏 "...that simplified is + . So then I added those two fractions together and got .." That shouldn't make any sense and if I saw your paper I may understand why you did it. But it's simply wrong. 1+ a b 1b a + 1b b b a + You are right so far so let's add these. b b b+a ... b and I don't know why you added a + b to get ab (probably a misplaced mark on the scratch paper). The b+a b final answer is ===== " its 𝑎+𝑏 𝑏 which is a+1 when simplified". Oooh, no. Both the a and the b are over the b. You can't cancel out just one. a+b b  a+ b b You can factor the b out of both but then you get a + b a / b +1 a = = +1 b 1 b but then you have worked yourself back to exactly where you started. Question #3 Laurent expansion of 1 z2 I need to find a Laurent expansion of If it was 1 z0 = 1P(1, 2014, 2015) z2 1 , I'd rewrite the fraction like this: z 1 ( z − 1) + 1 But with z 2 I'm not sure how to proceed... My attempt: According to the given recommendation, I've tried to do the expansion of 1 1 = , a0 = 1, q = −( z − 1) = 1 − z z 1 + ( z − 1) 1 : z  (1 − z) n = 1 + (1 − z ) + (1 − z )2 + ... n =0 And after differentiation:  0 + 1 + 2(1 − z ) + 3(1 − z ) + ... = n(1 − z )n−1 2 n =0 Answer: Another way is: 2  1 1− 1  z  = z 2 ( z − 1) 2 = (1 − z ) 2 + 2(1 − z )3 + 3(1 − z ) 4 + 4(1 − z )5 + (1 − z ) 2 = 1 + 2(1 − z ) + 3(1 − z ) 2 + 4(1 − z )3 + using the product of convergent power series. Question #4 Find the minimum of  cyc a with 2(b + c) As said in the title, I have to find the minimum of the following: a 2(b + c)  cyc with a +b+c  0 In my very last attempt, I tried to work it out using AM-GM: Since a 1 = 2(b + c) 2  cyc a 2a  a(b + c) a + b + c a 2(a + b + c)  = 2 2(b + c) a+b+c The problem is, to get the minimum value, a =b+c b =c+a c +a +b or a = b = c = 0 . This is wrong since a, b, c  0 And after some calculus I found the minimum value is 3 a=b=c. 2 So what mistake I have made? Answer: By AM-GM a 2a 2a =  = 2 2(b + c) cyc 2 a(b + c) cyc a + b + c  cyc + which is infimum because we can use c → 0 and b =c. Question #5 Solve the Logarithmic Equations for x, please. This one is an exponential equation that I can't figure out.. 7 x − 2 = 53− x These two are logarithmic equations that I'm also having trouble with.. ln 3 x − 6 = −2 1025 =5 7 + e4 x These ones really stumped me. Any explanations would be greatly appreciated! Thank you! Answer: Hint for #1: use Michael Hardy's suggestion. Hint for #2: exponentiate both sides, using the fact that e Hint for #3: bring the e 4x ln x =x term to the numerator, isolate it, then take the natural log, using the fact that ln e = x x Question #6 How to prove that x y +  x+ y y x I am trying to prove that x y +  x+ y y x I tried some manipulations, like multiplications in What am I missing? Answer: Let a  0b  0 , we have that x yx = x x , but I'm still stuck with that. 𝑎2 𝑏 2 𝑎3 + 𝑏 3 + ⇔ ≥𝑎+𝑏 𝑏 𝑎 𝑎𝑏 ⇔ 𝑎2 − 𝑎𝑏 + 𝑏 ≥ 𝑎𝑏 ⇔ (𝑎 − 𝑏)2 ≥ 0 Question #7 Does there exist a , b, c, d such that a+b+c+d is an integer? 4 Let a , b, c, d are different positive integers, such that a,b,c, nor d is a multiple of a different variable. Question: Does there exist a , b, c, d for a+b+c+d =I 4 such that I is an integer? Answer: One way is to find two primes that are 1 mod 4 and two primes which are 3 mod 4. For instance, 5,7,11,13. Question #8 Advanced GCSE Indices At school, we've been asked to simplify the following:  S 9/10   3/5  S  −4 I've given it a go but haven't been taught this yet. Could someone please explain the process which one would use? Answer: c  Sa  a −b c ac −bc  b  = (S ) = S S  Question #9 Ordering an even and odd fraction that are close We know that 1/ 4  5 /11  1/ 2 I did it this way from small to large: 1 3 3 = 4  3 12 5 11 1 6 6 = 2  6 12 It is hard for me to decide between 5 11 and 6 12 without a calculator. Is there an easier way to compare them? Answer: Multiply all the numbers by the least common multiple of the denominators - 44 , and you get the inequality: 11  20  22 Question #10 Simplifying Rational Expressions Simplify the following rational expression: 5 / ( x + 3) − 7 x / ( x − 1) I came across this question in my homework and because it is a fraction, I decided that I needed to establish a common denominator of: x −3 But how do I convert 𝑥 − 1 into 𝑥 − 3 ? Answer: Notice, we can simplify the given expression as follows 5 7x − x + 3 x −1 = 5 7x − 7 + 7 − x+3 x −1 = 5 7( x − 1) + 7 − x+3 x −1 = 5 7( x − 1) 7 − − x+3 x −1 x −1 = 5 7 −7− x+3 x −1 = −7 + 5 7 − x + 3 x −1 Or Using partial fraction 5 7x − x + 3 x −1 = 5( x − 1) − 7 x( x + 3) ( x + 3)( x − 1) = 5 x − 5 − 7 x 2 − 21x ( x + 3)( x − 1) = −7 x 2 − 16 x − 5 ( x + 3)( x − 1) = −(7 x 2 + 16 x + 5) ( x + 3)( x − 1) = −7 − 2 x + 26 ( x + 3)( x − 1) = −7 − 2( x + 13) ( x − 1)( x − 3) 𝑥+13 Making partial fractions of (𝑥−1)(𝑥−3) = 𝐴 𝑥−1 + 𝐵 𝑥+3 7 5 2 2 ⇒ 𝐴 = , 𝐵 = − , we get 5 7x 7 5 − = −7 − + x + 3 x −1 x −1 x + 3 Question #11 Simplify the expression: 32 n − 1 3n+1 − 3 Factorizing indices is quite a cool and intriguing area of indices. However, I stumbled upon one that I couldn't do any further: 32 n − 1 3n+1 − 3 I got up to this stage: 3n  3n − 30 3n  31 − 31 I got to this stage through using the laws of indices and breaking them down. I feel that I'm close, but not there yet. Please help. Thank you! Answer: ( ) ( )( ) ( ) ( ) 2 3n − 12 3n − 1 3n + 1 3n + 1 32 n − 1 = = = 3n+1 − 3 3 3n − 1 3 3 3n − 1 Question #12 Decompose a fraction in a sum of two Let's say that I have this fraction: 2x x + 4x + 3 2 I would like to decompose in two fraction: A B + x + 3 x +1 Which is the procedure for that? :) Answer: 2x 2x 2x = 2 = = x + 4 x + 3 x + x + 3x + 3 x( x + 1) + 3( x + 1) 2 = 2x A B = + ( x + 3)( x + 1) x + 3 x + 1 2 x = Ax + A + Bx + 3B = ( A + B ) x + A + 3B You needs to solve the system A + B = 2, A + 3B = 0 Question #13 Conditions present after solving for x The question asked me to solve x2 + 2 x − 8 =3 x2 − x − 2 The answer is x = 0.5x  2 . To get full marks. Why is this condition present? How can I recognise when a condition exists, when I encounter similar problems? Answer: When solving this equation you assume that x − x − 2  0  x  −1, 2 . 2 In general while solving an equation make sure that the denominator is not equal 0 . Question #14 If 1 1 1 + + = 1 ab + ac + bc  3 2+a 2+b 2+c Answer: The condition gives: 3 1  1 −1 =   −  2 2+a  cyc  2 1=  cyc a ( a + b + c )2  2+a a+b+c+6  3  ab + bc + ca by Cauchy-Schwarz inequality Question #15 How to differentiate for critical points with variable in denominator sorry for posting a particular problem, this is maybe more of an algebra problem than a calculus problem, but it does involve differentiating so I thought I would state the problem as one. I am currently running through the openOnlineMIT multivariable calculus course. I have ran into a problem which I can't seem to come to the solution. Specifically I am trying to find When I set this to zero I get d 4 =0 2 dt 1 + t + (1.5 − t 2 ) 2 1 = 04t (t 2 − 1) = 0 . which is a bit different but is solvable. t (1 − t 2 ) Answer: We have  d 4 16t 3 − 16t = −   dt  1 + t 2 + (1.5 − t 2 )2  (t 4 − 2t 2 + 3.25)2 This is zero only when 16t 3 − 16t = 0 16t (t 2 − 1) = 0 Which is basically the answer you were given. Did you make a mistake in taking the derivative? Question #16 Property of fractions   h+h hh Given two fractions both in reduced form. I am unable to find a case when  does not lie in  kk k +k  h h the interval [ ,  ] . Is there such a case ? k k Answer: Let h h =a =b k k Then, h + h ak + bk  = k + k k + k WLOG, let ab ak + bk  ak + ak   =a k + k k + k ak + bk  bk + bk   =b k + k k + k a ak + bk  b k + k a ak + bk  b k + k h h + h h   k k + k k The case where a = b . and where a  b is left to the reader as an exercise. Question #17 Intuitively, why does a 1 = ? c c a For intuition, I reference objects. Imagine making a dessert with: ac as chestnuts. Question. How and why is a c (3) = 1 c a (4) ? I ask only about intuition; please omit formal arguments and proofs (eg: Intuition is not generated by the explanation that rationalising (4)'s denominator produces (3)). Answer: The fraction ac is the number of chestnuts per apple. ca Suppose that you have x chestnuts per apple, where x can be any positive real number. No matter what xx chestnuts must be getting 1 1 of an apple, so there are apples per chestnut. x x Thus, in the particular case that x = c , it must be the case that a a 1 = . c c a Question #18 Partial fractions - alternative result I have the following fraction: z ( z − 1)( z − 2) When I try to decompose it to partial fractions, I get: − 1 2 + z −1 z − 2 But the result in my book is: − z z + z −1 z − 2 Both results are correct, but, how am I supposed to get the second one? Answer: I agree with the comments that the result stated in the back of the book is somewhat non-standard. The way that you can see that the two results are the same is by cleverly adding and subtracting 1. −1 2 + z −1 z − 2 −1 2 −1+ +1 z −1 z−2 −1 ( z − 1) 2 z−2 = − + + z −1 z −1 z − 2 z − 2 −z z = + z −1 z − 2 = Question #19 1 1 1 + + + 11 12 200  19 Prove that 1 1 1 + + + 10 11 1112 19  20 My attempt:The denominator is 1 1 + + 10 11 11 12 + 1 using telescopic series. 20 1 1 1 1 1 = − + − + 19  20 10 11 11 12 − 1 1 1 1 = − = 20 10 20 20 I think we could Prove that the top part of fraction is bigger than 1 but I don't know how. Answer: Your calculation for the denominator (telescopic sum and not series) was the hard part, since 1 1 1 + + + 11 12 200 1 20 1 1 + +  200 200 1 20 + 1 200 − 10 19 200 = 200 = 20 = 19 1 1 20 20 Question #20 Hard question simplify as much as possible following x3 − 49 x x 4 − 13x3 + 50 x 2 − 56 x I get answer x+3 but my friend say it's wrong I dont understand please help me please x −1 Answer: x 3 − 49 x ( x 4 − 13 x3 + 50 x 2 − 56 x) x( x 2 − 49) x( x3 − 13 x 2 + 50 x − 56) x( x − 7)( x + 7) = x( x − 2)( x − 4)( x − 7) =