Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
History of Grandi's series wikipedia , lookup
Line (geometry) wikipedia , lookup
Numerical continuation wikipedia , lookup
System of polynomial equations wikipedia , lookup
Recurrence relation wikipedia , lookup
System of linear equations wikipedia , lookup
§§4.2-4.4 Review Problems MTH 111 - Spring 2014 Friday, May Name: 2nd Final Exam: 7:00-10:00pm Score: /10 This short review contains questions from Sections 4.2 - 4.4 (which were not previously covered on a past exam). The questions included in this review are by no means exhaustive. This review simply serves to provide extra practice on topics more recently discussed in your MTH 111 course. The final exam is cumulative, so you should consult your old exams and the past exam reviews to study for the final. You should also continue to read the textbook, consult your instructor, and work on the suggested problems listed on the syllabus for more exam preparation. Good luck! 1 1. Evaluate log4 (64). Solution: Here we want to find the power of 4 that yields 64. Since 43 = 64, we conclude the above logarithm is equal to 3; algebraically, we have that log4 (64) = x ⇐⇒ 4x = 64 = 43 =⇒ log4 (64) = 3. 1 1 2. Evaluate log3 ( 27 ). Solution: 3 raised to the −3 power yields log3 1 1 27 1 27 : = x ⇐⇒ 3x = 1 1 = 3 = 3−3 =⇒ log3 27 3 1 27 = −3 3. Evaluate ln(1). Solution: Recall that loga (1) = 0 for any a > 0, a 6= 1. Equivalently, ln(1) = x ⇐⇒ ex = 1 = e0 ⇐⇒ ln(1) = 0. 1 4. Find the domain of the function f (x) = log3 (5x − 7). Solution: Since we can only consider the logarithms of strictly positive numbers, we must stipulate that the interior of this function is always greater than zero. Since 5x − 7 > 0 implies that 5x > 7 or x > 57 , we have that the domain of the function is ( 75 , ∞). 1 5. Given f (x) = 13 · 5x−4 , find the inverse f −1 . Solution: Write y = 31 · 5x−4 . First, switch x and y: x= 1 y−4 ·5 . 3 Multiply both sides by 3: 3x = 5y−4 . By the definition of a logarithm, we can convert this equation to one of logarithmic form: log5 (3x) = y − 4. Finally, add 4 to both sides to yield the following: y = log5 (3x) + 4. Thus f −1 (x) = log5 (3x) + 4. 1 6. Solve log7 (x) = 2. Solution: Convert this equation to one of exponential form: log7 (x) = 2 ⇐⇒ x = 72 = 49, so x = 49. 1 7. Solve ln(x2 ) = ln(2x). Solution: Recall the following important property of logarithms: One-to-One Property of Logarithms For a > 0 and a 6= 1, if loga (x) = loga (y) then x = y. Since ln(x2 ) = ln(2x), we may conclude x2 = 2x. But then x2 = 2x ⇐⇒ x2 − 2x = 0 ⇐⇒ x(x − 2) = 0 =⇒ x = 0 or x = 2. However, we cannot plug x = 0 into the original logarithmic equation since we cannot find the logarithm of 0. Thus this is an extraneous solution, and our only true solution to the original equation is x = 2. 1 8. Rewrite the expression log6 (3) + 14 log6 (x) − 2 log6 (y) − 3 log6 (z) as a single logarithm. Solution: First use the Power Rule to bring “coefficients” of logarithms to powers of the interiors of the functions. Also, factor out −1 from the last two logarithms: log6 (3) + 1 1 log6 (x) − 2 log6 (y) − 3 log6 (z) = log6 (3) + log6 (x 4 ) − log6 (y 2 ) + log6 (z 3 ) . 4 Using the Product Property, we have that 1 1 log6 (3) + log6 (x 4 ) − log6 (y 2 ) + log6 (z 3 ) = log6 (3x 4 ) − log6 (y 2 z 3 ), but this difference can be rewritten with the Quotient Property: 1 4 2 3 log6 (3x ) − log6 (y z ) = log6 1 3x 4 . y2z3 1 √ 73a 9. Rewrite the expression ln as a sum or difference of multiples of logarithms. b2 Solution: Rewrite the quotient with a difference of logarithms. Then apply the Product and Power Rules. √ √ 1 73a 1 = ln(7 3 a) − ln(b2 ) = ln(7) + ln(a 3 ) − ln(b2 ) = ln(7) + ln(a) − 2 ln(b) ln b2 3 1 10. Solve log7 (x2 ) = log7 (6 − x). Solution: Apply the One-to-One Property to the above equation to conclude that x2 = 6 − x. Rewrite this equation and factor! x2 = 6 − x ⇐⇒ x2 + x − 6 = 0 ⇐⇒ (x + 3)(x − 2) = 0 =⇒ x = −3, or x = 2. Although one of our solutions is negative, we will show that neither solution is extraneous. Plug each potential solution into the original equation to see if the equality is indeed satisfied. ? X 1) x = 2 =⇒ log7 (22 ) = log7 (6 − 2) ⇐⇒ log7 (4) = log7 (4) ? X 2) x = −3 =⇒ log7 ((−3)2 ) = log7 (6 − (−3)) ⇐⇒ log7 (9) = log7 (9) Since each equation is properly satisfied, are solutions are indeed x = −3 and x = 2. Good luck on the final exam! ,