Algebra/Trig Review - Pauls Online Math Notes - Lamar University

More documents

Recommendations

Info

$Ordinary Differential Equations, ODE, Math 3301 ... - Lamar University$

$Common Derivatives and Integrals - Pauls Online Math Notes$

$Calculus and Analytic Geometry III, Math 2415 ... - Lamar University$

$Algebra Cheat Sheet ... - Pauls Online Math Notes$

$Introduction to Linear Algebra, MATH 2318/3328 - Lamar University$

$How to Study Mathematics - Pauls Online Math Notes - Lamar ...$

Algebra/Trig ReviewSo, the two solutions are x = 0 and2x = − .5Solving Logarithm EquationsSolving logarithm equations are similar to exponential equations. First, we isolate thelogarithm on one side by itself with a coefficient of one. Then we use Property 4 fromthe Logarithm Properties section with an appropriate choice of b. In choosing theappropriate b, we need to remember that the b MUST match the base on the logarithm!Solve each of the following equations.1. ( x)4log 1− 5 = 2Solution© 2006 Paul Dawkins 94http://tutorial.math.lamar.edu/terms.aspx
Algebra/Trig ReviewThe first step is to divide by the 4, then we’ll convert to an exponential equation.1log ( 1− 5x)=21log( 1−5 x)210 = 101− 5x= 10Note that since we had a common log in the original equation we were forced to use abase of 10 in the exponential equation. Once we’ve used Property 4 to simplify theequation we’ve got an equation that can be solved.121− 5x= 10− 5x=− 1+10121211 ⎛ ⎞2x =− ⎜− 1+10 ⎟5 ⎝ ⎠x = -0.4324555320Now, with exponential equations we were done at this point, but we’ve got a littlemore work to do in this case. Recall the answer to the domain of a logarithm (theanswer to Problem 9 in the Logarithm Properties section). We can’t take thelogarithm of a negative number or zero.This does not mean that x = − 0.4324555320 can’t be a solution just because it’snegative number! The question we’ve got to ask is this : does this solution produce anegative number (or zero) when we plug it into the logarithms in the originalequation. In other words, is 1− 5x negative or zero if we plug x = − 0.4324555320into it? Clearly, (I hope…) 1− 5x will be positive when we plug x = − 0.4324555320in.Therefore the solution to this is x = − 0.4324555320.Note that it is possible for logarithm equations to have no solutions, so if that shouldhappen don’t get to excited!⎛ x ⎞2. 3 + 2ln ⎜ + 3⎟=−4⎝7⎠SolutionThere’s a little more simplification work to do initially this time, but it’s not too bad.© 2006 Paul Dawkins 95http://tutorial.math.lamar.edu/terms.aspx
Page 1 and 2:
Algebra/Trig ReviewIntroductionThis
Page 3 and 4:
Algebra/Trig ReviewSimplifying Rati
Page 5 and 6:
Algebra/Trig Review1 1 1 13− −
Page 7 and 8:
Algebra/Trig Review⎧33 −8xif x
Page 9 and 10:
Algebra/Trig ReviewTo convert radic
Page 11 and 12:
Algebra/Trig ReviewSo, what do we d
Page 13 and 14:
Algebra/Trig Review( )2f x =− x +
Page 15 and 16:
Algebra/Trig Review⎛ f ⎞⎜ ⎟
Page 17 and 18:
Algebra/Trig ReviewYou can memorize
Page 19 and 20:
Algebra/Trig Review6.SolutionIn thi
Page 21 and 22:
Algebra/Trig Review2. ( ) 2y = x+ 3
Page 23 and 24:
Algebra/Trig ReviewWe could also us
Page 25 and 26:
Algebra/Trig Review26. ( ) 2x+ y+ 5
Page 27 and 28:
Algebra/Trig Review2 29. ( x+ 1 ) (
Page 29 and 30:
Algebra/Trig Review12. y = xSolutio
Page 31 and 32:
Algebra/Trig ReviewComplex numbers
Page 33 and 34:
Algebra/Trig ReviewFrom this we see
Page 35 and 36:
Algebra/Trig ReviewSo, now that we
Page 37 and 38:
Algebra/Trig ReviewSo, the two inte
Page 39 and 40:
Algebra/Trig Review3.23x−2x− 11
Page 41 and 42:
Algebra/Trig ReviewIn this case we
Page 43 and 44: Algebra/Trig ReviewSolutionThis one
Page 45 and 46: Algebra/Trig Review4x+ 534x−2x
Page 47 and 48: Algebra/Trig ReviewRemember how the
Page 49 and 50: Algebra/Trig ReviewFrom this unit c
Page 51 and 52: Algebra/Trig ReviewNow, if we remem
Page 53 and 54: Algebra/Trig ReviewTo do this probl
Page 55 and 56: Algebra/Trig ReviewIf ω > 1we can
Page 57 and 58: Algebra/Trig ReviewIn the case of t
Page 59 and 60: Algebra/Trig ReviewSolutionCotangen
Page 61 and 62: Algebra/Trig Review2x = 1 ( + ( x))
Page 63 and 64: Algebra/Trig Reviewdetermine what t
Page 65 and 66: Algebra/Trig Review3Now, there are
Page 67 and 68: Algebra/Trig Review4π2πn− 4x= +
Page 69 and 70: Algebra/Trig ReviewSince we want th
Page 71 and 72: Algebra/Trig Reviewπw= + 2 π n, n
Page 73 and 74: Algebra/Trig ReviewI didn’t put i
Page 75 and 76: Algebra/Trig ReviewSo, using our ca
Page 77 and 78: Algebra/Trig ReviewThere’s anothe
Page 79 and 80: Algebra/Trig ReviewFrom this we can
Page 81 and 82: Algebra/Trig Review5.⎛cos cos⎜
Page 83 and 84: Algebra/Trig Reviewx3. List as some
Page 85 and 86: Algebra/Trig ReviewBasic Logarithmi
Page 87 and 88: Algebra/Trig ReviewSolutionlog b xb
Page 89 and 90: Algebra/Trig ReviewThe domain y = l
Page 91 and 92: Algebra/Trig ReviewIn each of the e
Page 93: Algebra/Trig Review1−3z7 + 15e= 1
Page 97 and 98: Algebra/Trig Reviewx 2= e1−x2x= e
show all

Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Create successful ePaper yourself

Delete template?

Save as template?