Questions and Answers #9 Logarithms

Questions and Answers Sheet 9 Logarithms Question #1 Proving an inequality without an integral: 1 𝑥+1 ≤ ln(1 + 𝑥) − ln(𝑥) ≤ 1 𝑥 Answer: Let 𝑓(𝑥) ≔ ln(𝑥), then the Mean Value Theorem (Differentiation) says that there exists some 𝜉 ∈ (𝑥, 𝑥 + 1) ln(1 + 𝑥) − ln(𝑥) = As 𝑑𝑓 𝑑𝑥 (𝑥) = 1 ln(1+𝑥)−ln(𝑥) (1+𝑥)−𝑥 = 𝑓(1+𝑥)−𝑓(𝑥) and as the function 𝑥 (1+𝑥)−𝑥 1 𝑥 = 𝑑𝑓 𝑑𝑥 𝜉 is monotonic we know that Question #2 1 ∞ log(𝑥+𝑥) How to solve ∫0 𝑑𝑥 ? 1+𝑥 2 Answer: You may write 1 ∞ log(1+𝑥 2 )−log 𝑥 𝑑𝑥 = ∫0 𝑑𝑥 1+𝑥 2 1+𝑥 2 ∞ log(1+𝑥 2 ) ∞ log 𝑥 ∫0 1+𝑥 2 𝑑𝑥 − ∫0 1+𝑥 2 𝑑𝑥 ∞ log(𝑥+𝑥) ∫0 = Clearly, by the change of variable 1 → , we get 𝑥 ∞ log 𝑥 = ∫0 1+𝑥 2 ∞ log 𝑥 𝑑𝑥 = − ∫0 1+𝑥 2 𝑑𝑥 On the other hand, by the change of variable 𝑥 = tan 𝜃 , 𝑑𝑥 = (1 − tan2 𝜃)𝑑𝜃, 1 + tan2 𝜃 = 1 cos 2 𝜃 we obtain the classic evaluation: 𝜋 log(1 + 𝑥 2 ) 2 =∫ 𝑑𝑥 = −2 ∫ log(cos 𝜃) 𝑑𝜃 = 𝜋 log 2 2 1 + 𝑥 0 0 ∞ Question #3 Given log 𝑥 𝑏−𝑐 = log 𝑦 𝑐−𝑎 = log 𝑧 𝑎−𝑏 show that 𝑥 𝑏+𝑐−𝑎 ⋅ 𝑦 𝑐+𝑎−𝑏 ⋅ 𝑧 𝑎+𝑏−𝑐 = 1 Answer: We have log 𝑥 𝑏−𝑐 = log 𝑦 𝑐−𝑎 = log 𝑧 𝑎−𝑏 =𝑡 This gives us 𝑥 = 𝑒 𝑡(𝑏−𝑐) , 𝑦 = 𝑒 𝑡(𝑐−𝑎) and 𝑧 = 𝑒 𝑡(𝑎−𝑏) Hence, 𝑥 𝑏+𝑐−𝑎 ⋅ 𝑦 𝑐+𝑎−𝑏 ⋅ 𝑧 𝑎+𝑏+𝑐 = 𝑒 ((𝑏−𝑐)(𝑏+𝑐−𝑎)+(𝑐−𝑎)(𝑐+𝑎−𝑏)+(𝑎−𝑏)(𝑎+𝑏−𝑐)) 2 2 2 2 2 2 = 𝑒 𝑡(𝑏 −𝑐 −𝑎𝑏+𝑎𝑐+𝑐 −𝑎 −𝑏𝑐+𝑏𝑎+𝑎 −𝑏 −𝑎𝑐+𝑏𝑐) = 𝑒 0 = 1 Question #4 1 𝑥+1 ≤ 𝑑𝑓 𝑑𝑥 (𝜉) ≤ 1 𝑥 1 Ways of showing ∑∞ 𝑛=1 ln (1 + ) to be divergent 𝑛 Answer: Notice the following: 1 𝑛+1 𝑛 𝑛 log (1 + ) = log ( ) = log(𝑛 + 1) − log(𝑛) Hence 1 ∑𝑛𝑘=1 log (1 + ) = log(𝑛 + 1) → ∞ 𝑘 Question #5 How do I convince students in high school for which this equation: 2𝑥 = 4𝑥 have only one solution in integers that is x=4? Answer: Hint: plot the graph of 𝑦 = 2𝑥 and y=4x and shows that the only other solution is between 0 and 1. Question #6 Find a branch of 𝑓(𝑧) = log(𝑧 3 − 2) that is analytic at z=0 . Answer: Note that, z=0 is not a branch point of f(z). To find the branch points of f(z), solve the equation 1 2𝑘𝜋𝑖 3 ,𝑘 𝑧 3 − 2 = 0 ⇒ 𝑧 3 = 2𝑒 2𝑘𝜋𝑖 ⇒ 𝑧 = 23 𝑒 = 0,1,2 Question #7 Proving 𝑖 2 ln 𝑥+𝑖 𝑥−𝑖 = arctan 𝑥 Answer: HINT: Let arctan 𝑥 = 𝑦 ⇒ 𝑥 = tan 𝑦 𝑥+𝑖 𝑥−𝑖 = sin 𝑦+𝑖 cos 𝑦 sin 𝑦−𝑖 cos 𝑦 = 𝑖(cos 𝑦−𝑖 sin 𝑦) −𝑖(cos 𝑦+𝑖 sin 𝑦) = −𝑒 −2𝑖𝑦 Question #8 Prove that | log(1 + x ) − log(1 + y ) || x − y | 2 2 Answer: The derivative of 𝑓(𝑥) = log(1 + 𝑥 2 ) is 2𝑥 𝑓 ′ (𝑥) = 2 1+𝑥 and | 2x | 1 1 + x2 because 2|𝑥| ≤ 1 + |𝑥|2 since (1 − |𝑥|)2 ≥ 0 Thus, by the mean value theorem, log(1+𝑥 2 )−log(1+𝑦 2 ) 𝑥−𝑦 = 𝑓 ′ (𝑧) for z between x and y (assuming 𝑥 ≠ 𝑦 or the inequality is obvious). Then | log(1 + x 2 ) − log(1 + y 2 ) |=| f ( z )  1| x− y Question #9 Show that log(𝑥 + 1) − log(𝑥) < 1 for x>0 𝑥 Answer: The function log(1 + 𝑡) is strictly concave and therefore its graph stays under its tangent line at 0: for any 𝑡 ≠ 0 and t>-1, log(1 + 𝑡) < 𝑡. Your inequality is equivalent to 1 1 𝑥 𝑥 log(𝑥 + 1) − log(𝑥) = log (1 + ) < Question #10 How to show that log ( 2𝑎 1−𝑎2 + 2𝑏 1−𝑏 2 + 2𝑐 1−𝑐 2 ) = log 2𝑎 1−𝑎2 + log 2𝑏 1−𝑏 2 + log 2𝑐 1−𝑐 2 Answer: HINT: Use log 𝑎 + log 𝑏 + log 𝑐 = log(𝑎𝑏𝑐) and then put 𝑎 = tan 𝐴 etc. to find ∑ tan 𝐴 = ∏ tan 𝐴 ⇒ 𝐴 + 𝐵 + 𝐶 = 𝑛 𝜋 where n is any integer and 2𝑎 1−𝑎2 = ⇒ ∑ tan 2 𝐴 = ∏ tan 2 𝐴 as 2𝐴 + 2𝐵 + 2𝐶 = 2𝑛 Now apply logarithm now Question #11 𝑋 = log12 18 and 𝑌 = log 24 54. Find XY+5(X-Y) Answer: Let 𝐼 = log 18 log 12 ⋅ log 54 log 24 + 5( log 18 log 12 − log 54 log 24 ). Also, let log 3 = 𝑥 and log 2 = 𝑦. Then, 𝐼= = = log 32 ⋅2 log 22 ⋅3 ⋅ log 33 ⋅2 log 23 ⋅3 +5( log 32 ⋅2 log 22 ⋅3 − log 33 ⋅2 )= log 23 ⋅3 6𝑥 2 +5𝑥𝑦+𝑦 2 +10𝑥 2 +35𝑥𝑦+15𝑦 2 −35𝑥𝑦−10𝑦 2 2𝑥+𝑦 2𝑦+𝑥 ⋅ 3𝑥+𝑦 3𝑦+𝑥 + 5( 2𝑥+𝑦 2𝑦+𝑥 − (2𝑦+𝑥)(3𝑦+𝑥) 𝑥 2 +5𝑥𝑦+6𝑦 2 𝑥 2 +5𝑥𝑦+6𝑦 2 =1 Question #12 Prove (1 − 𝑥) ln(1 − 𝑥) + (1 + 𝑥) ln(1 + 𝑥) ≤ 2𝑥 2 for 0 < 𝑥 < 1 Answer: We have that ⋆ ln(1 + 𝑥) ≤ 𝑥 3𝑥+𝑦 3𝑦+𝑥 ) 2 tan 𝐴 1−tan2 𝐴 = tan 2 𝐴 ⋆ ln(1 − 𝑥) ≤ −𝑥 therefore (1 − 𝑥) ln(1 − 𝑥) + (1 + 𝑥) ln(1 + 𝑥) ≤ −𝑥 + 𝑥 2 + 𝑥 + 𝑥 2 = 2𝑥 2 Question #13 log(𝑛) is what power of n? Answer: No: If ∈ is any positive number, then 𝑛𝜖 grows faster than log 𝑛. This can be made precise in the log 𝑛 statement lim 𝑛→∞ 𝑛𝜖 = 0 for all 𝜖 > 0. To prove this, just note that by L'Hospital's rule, 1 log 𝑛 1 1 𝑛 lim = lim = lim 𝜖 = 0 𝑛→∞ 𝑛𝜖 𝑛→∞ 𝜖𝑛𝜖−1 𝜖 𝑛→∞ 𝑛 Question #14 How can I deduce that without Taylor series or L'Hospital's rule? lim ln(1+𝑥) 𝑥 𝑥→0 =1 Answer: If the limit exists, say L, then you can state that: 𝐿 = lim ln(1+𝑥) 𝑥 𝑥→0 lim 𝐿 ln(1+𝑥) 𝑥 ∴ 𝑒 = 𝑒 𝑥→0 = lim 𝑒 ln(1+𝑥) 𝑥 𝑥→0 1 = lim(𝑒 ln(1+𝑥) )𝑥 𝑥→0 1 = lim(1 + 𝑥)𝑥 𝑥→0 =𝑒 ∴𝐿=1 Question #15 Prove that: lim ln(cos 𝑥) 2 𝑥→0 ln(1−𝑥 ) =1 2 without L'Hopital's rule. Answer: Using ln(1 − 𝑦) = − ∑∞ 𝑛=1 ln (1 − 𝑥2 2 )=− cos 𝑥 = 1 − Therefore 𝑥2 𝑥2 2 𝑦𝑛 𝑛 we get immediately + 𝑂(𝑥 4 ), and also, after observing that + 𝑂(𝑥 4 ), ln cos 𝑥 = − 2 ln(cos 𝑥) 𝑥2 ) 2 ln(1− = 1 + 𝑂(𝑥 2 ) Question #16 𝑥2 2 + 𝑂(𝑥 4 ). Limit of function with natural logarithm I need help solving this problem. I tried L'hospital and rearranging but nothing worked. 1 1 𝑥 𝑥+1 lim (ln (1 + ) − 𝑥→∞ ) Answer: Using L'Hospital rule we get 𝑥 1 1 1 1 ⋅ (− 2 ) + ln (1 + ) − 𝑥 + 1 (𝑥 𝑥 + 1)2 𝑥 𝑥 + 1 = lim lim = 1 −2 𝑥→∞ 𝑥→∞ 2 3 𝑥 𝑥 −𝑥−1+𝑥 𝑥 3 1 𝑥2 1 lim = lim (𝑥+1)2 ⋅ = 2 ⋅ 𝑥→∞ 𝑥(𝑥+1) 2 2 𝑥→∞ 2 Question #17 How to find the value of an unknown exponent? E.g. I have the question: 24𝑥+1 = 128 I solved this by knowing that 128 = 27 and therefore x must equal 1.5. However, is there a way of solving this without knowing that 128 = 27 ? Answer: ... and without logarithms or knowing any powers of 2 other than the most trivial one ... 24𝑥+1 = 128 24𝑥+1−1 = 24𝑥 = 4𝑥−1 128 2 = 64 2 = 32 24𝑥−2 = 16 24𝑥−3 = 8 24𝑥−4 = 4 24𝑥−5 = 2 24𝑥−6 = 1 = 20 6 3 so 4x-6=0 and 𝑥 = = . 4 2 Question #18 Is there a different approach to evaluate ∫ ln(𝑥) 𝑑𝑥? Answer: ∞ 1 ln 𝑥 = ∑(−1)𝑛+1 (𝑥 − 1)𝑛 𝑛 𝑛=1 ∞ ∫ ln 𝑥 𝑑𝑥 = ∑(−1)𝑛+1 𝑛=1 1 𝑛+1 = ∑∞ ( − 𝑛=1(−1) 𝑛 = (𝑥 − 1 1 1 (𝑥 − 1)𝑛+1 ⋅ 𝑛 1+𝑛 ) (𝑥 − 1)𝑛+1 𝑛+1 1 ∞ 𝑛+1 (𝑥 1) ∑𝑛=1(−1) 𝑛 𝑛+2 − 1)𝑛 + ∑∞ 𝑛=1(−1) = (𝑥 − 1) ln 𝑥 + ln 𝑥 − (𝑥 − 1) = 𝑥 ln 𝑥 − 𝑥 + 𝐶 1 𝑛+1 (𝑥 − 1)𝑛+1 Question #19 Prove ∑∞ 𝑛=2 1 is divergent. 𝑛 ln(𝑛) ln ln 𝑛 Answer: Your series diverges by the integral test, because 1 𝑑𝑥 = log(log(log(𝑥))) ∫ 𝑥 log(𝑥) log(log(𝑥)) and because lim log(log(log(𝑥))) = +∞ 𝑥→+∞ Concerning your approach, note that the inequality 1 𝑛 ln2 𝑛 < 1 𝑛2 is false. Actually, 𝑛 ln2 𝑛 < 𝑛2 . Question #20 8 5 47 54 Prove log 1 > log 1 Answer: Here is another argument using power-series-based computations of log, but slightly different in its details to Ivan's. 8 8 7 1 47 7 8 8 log 1 = − log 4 = log 4 = log 4 (1 − ), while 5 5 4 1 log 1 = − log 5 = log 5 = log 5 (1 − ) 4 4 5 5 5 So the problem reduces to showing that 1 1 8 5 log 4 (1 − ) > log 5 (1 − ). This is slightly delicate, because both numbers of which we are taking the 𝑙𝑜𝑔 are close to 1, so both sides are close to 0, and we have to estimate how close. Recall that ln(1 − 𝑥) = −𝑥 − 𝑥2 2 − 𝑥3 3 − ⋯ 1 when |x|<1, and that log 𝑎 (1 − 𝑥) = ( ) ln(1 − 𝑥). So we have to prove ln 𝑎 1 1 1 1 ln 4 ( + + ⋯ ) > ln 5 ( + + ⋯ ). 5 5 8 128 Now clearly 1 1 5 50 ( + 8 1 1 5 8 128 + ⋯) > ( + + ⋯ ), so it suffices to show that 8 ( ) ln 4 > ln 5, 5 i.e. that ln 48 > ln 55 , i.e. that 65536>3125, which is true.