Section 5.6 logarithmic and exponential equations
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This document contains solutions to several logarithmic and exponential equations. It begins by solving the equations log 4 2log x= and ( ) ( )2 2 1x x+ − =, finding the values of x that satisfy each. It then solves equations involving logarithms and exponents such as 3 7x = ln3 ln 7x, 5 2 3x × = 3, and 1 2 3 ln 2 ln5x x− + =. Each solution provides the step-by-step work and resulting value of x. The document concludes by solving the quadratic equation 9 3 6 0x x − − =.
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Rational expressions involve fractions of polynomials. The domain of an algebraic expression is the set of real numbers for which the expression is defined. Compound fractions contain fractions in the numerator, denominator, or both. Rational expressions can be simplified by factoring and canceling common factors. Adding and subtracting rational expressions requires finding a common denominator. The denominator of a fraction can be rationalized by multiplying the numerator and denominator by the conjugate radical. Common errors involve applying properties of multiplication to addition incorrectly.
Polynomial Function and Synthetic Division
This presentation explains the basic information about Polynomial Function and Synthetic Division. Examples were given about easy ways to divide polynomial function using synthetic division. It also contains the steps on how to perform the division method of polynomial functions.
indice-ppt.ppt
1. Indices, surds, logarithms and exponential functions are covered. Key points include laws of indices, rationalizing surds, laws of logarithms including change of base, and the general forms of exponential functions as ax and ex.
2. Example questions demonstrate working with indices, simplifying surds, solving logarithmic and exponential equations using appropriate logarithm laws and change of variable techniques.
3. Solutions to example equations involve algebraic manipulation and setting logarithmic and exponential terms equal to solve for variables.
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R.5-day2-Multiplying-and-Dividing-Rational-Expressions.ppt
R.5-day2-Multiplying-and-Dividing-Rational-Expressions.ppt
Section 13.6 solving quadratic equations using the zero-factor property
Section 13.6 solving quadratic equations using the zero-factor property
Section 14.6 solving equations with rational expressions
Section 14.6 solving equations with rational expressions
Special topics about stocks and bonds using algebra
Special topics about stocks and bonds using algebra
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Natural rubber
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Fat and oil
Oils and fats are esters composed of glycerol and fatty acids. Fats are found in animals and are solids at room temperature, while oils can be found in animals and plants and are liquids at room temperature. The document defines saturated and unsaturated fatty acids and lists examples of each found in common fats like palm oil. It describes the differences between saturated and unsaturated fats and their properties. Finally, it discusses some advantages of palm oil including its widespread use in foods and importance to the economy through business and jobs.
Ester
Esters are formed through an esterification reaction when a carboxylic acid reacts with an alcohol in the presence of a concentrated sulfuric acid catalyst. Esters have various physical properties including being colorless liquids at room temperature with sweet, pleasant smells and low boiling points and densities. They are found naturally in many plants and fruits where they contribute to smells and flavors. Esters have several applications including use as food flavorings, in cosmetics, fragrances, and medicines.
Carboxylic acid
Carboxylic acids have the general formula R-COOH. They are weak acids that only partially dissociate in water. Common properties include being colorless liquids or solids with sharp odors and high boiling points. Alcohols can be oxidized to form carboxylic acids using potassium dichromate and sulfuric acid. Carboxylic acids react with metals to form salts and hydrogen gas, with carbonates to form salts, carbon dioxide and water, and with bases to form salts and water. They also undergo esterification reactions with alcohols to form esters and water. Common uses include as preservatives and flavorings in food and in making soaps, drugs, dyes,
Alcohol
Alcohols are compounds containing a hydroxyl (-OH) group. They are named based on the carbon chain and position of the hydroxyl group. Alcohols can be produced through fermentation of sugars by yeast or through hydration of alkenes with steam. They have low boiling points, are colorless and volatile. Alcohols can undergo combustion, oxidation, and dehydration reactions. Ethanol is used as a fuel and solvent, while alcohols in general have industrial and medical uses.
Alkanes
Alkanes are saturated hydrocarbons whose general formula is CnH2n+2. Their names are derived from their molecular formula. Structural formulas show how atoms are bonded. Physical properties of alkanes include being soluble in organic solvents but not water, and existing as gases at low carbon numbers and liquids or solids at higher numbers. Melting and boiling points increase with more carbon atoms as intermolecular forces strengthen. Alkanes undergo combustion and halogenation reactions. Complete combustion produces CO2 and H2O while incomplete produces CO and H2O. Halogenation is a substitution reaction that occurs in sunlight, breaking C-H bonds and forming C-X bonds to produce chlorometh
Alkene
Alkenes are hydrocarbons containing at least one carbon-carbon double bond. They have lower melting and boiling points than alkanes due to weaker intermolecular forces. The number of carbons determines an alkene's name and formula. Alkenes undergo addition reactions, combustion reactions, polymerization reactions, and can be used to test for double bonds. They differ from alkanes in bonding, reactivity and ability to cause soot during combustion. Isomers are compounds with the same molecular formula but different structural formulas, resulting in different physical but same chemical properties.
Homologous series
A homologous series is a series of compounds with similar chemical properties where each member differs from the next by a CH2 group. The key characteristics of a homologous series are:
1) Each member can be represented by a common chemical formula that differs by CH2.
2) Members are prepared by a common method.
3) Members have the same chemical properties.
4) Each member differs from the next by one CH2 group which has a mass of 14.
Carbon compound
Carbon compounds can be divided into organic and inorganic compounds. Organic compounds contain carbon and are obtained from living things, having low boiling points. Inorganic compounds do not come from living things and have higher boiling points. Hydrocarbons are organic compounds made of only carbon and hydrogen. They can be saturated, containing only single bonds, or unsaturated, containing double or triple bonds. The molecular and structural formulas provide information on the atoms and bonds in a molecule. Naming carbon compounds according to IUPAC guidelines involves a stem/root indicating the number of carbons and an ending denoting the compound class.
Chapter 1 Rate of Reactions
The document discusses rate of reaction and factors that affect it. It defines rate of reaction as the change in amount of reactants or products per unit time. It describes several factors that affect rate based on collision theory, including surface area, concentration, temperature, catalysts, and pressure. It gives examples of how scientific understanding of rate of reaction enhances quality of life, such as refrigeration, pressure cooking, cutting food into smaller pieces, making margarine, and burning coal.
SPM F5 Chapter 1 Rate of Reaction
The document discusses rate of reaction and factors that affect it. It defines rate of reaction as the change in amount of reactants or products per unit time. Rate of reaction is affected by several factors including surface area, concentration, temperature, catalysts and pressure (for gas reactions). The collision theory is also explained, stating that reactions only occur during effective collisions where particles attain sufficient kinetic energy to overcome the activation energy barrier. Examples of how scientific understanding of rate of reaction enhances quality of life through applications like food storage, cooking and petroleum processing are provided.
Chapter 8 Alkyl halides
Halogenoalkanes, also known as alkyl halides, contain carbon-halogen bonds. They can be synthesized through free radical substitution or electrophilic addition reactions. Nucleophilic substitution reactions of halogenoalkanes produce alcohols or other products depending on the solvent. In aqueous solutions, hydroxide acts as a nucleophile to form alcohols via SN1 or SN2 mechanisms. In alcoholic solutions, hydroxide acts as a base to eliminate halogens and form alkenes. Both substitution and elimination reactions occur simultaneously but the solvent influences which pathway dominates.
Chapter 7 Alkenes and Alkyne
1) Alkenes are hydrocarbons that contain a carbon-carbon double bond. They include many naturally occurring compounds and important industrial materials.
2) The degree of unsaturation relates the molecular formula to possible structures by counting the number of multiple bonds or rings. Each double bond or ring replaces two hydrogens.
3) Alkenes react through electrophilic addition reactions, often involving a carbocation intermediate. The stability of the carbocation predicts the orientation of addition.
Chapter 05 an overview of organic reactions.
This document provides an overview of organic reactions, including the different types of organic reactions and how reaction mechanisms are used to describe the steps involved in organic reactions. It discusses several key aspects of organic reactions, including: 1) the common types of organic reactions such as addition, elimination, substitution, and rearrangement reactions, 2) how reaction mechanisms are used to describe the individual steps that occur in organic reactions, from reactants to products, and 3) the different types of steps that can be involved in reaction mechanisms, including the formation and breaking of covalent bonds. It also provides examples of reaction mechanisms, such as the addition of HBr to ethylene.
Chapter 05 stereochemistry at tetrahedral centers
This document discusses stereochemistry at tetrahedral carbons. It defines key terms like enantiomers, which are nonsuperimposable mirror images of each other. Enantiomers have different spatial arrangements but identical physical properties except for how they rotate plane-polarized light. The document also outlines Cahn-Ingold-Prelog rules for assigning R and S configurations to chiral centers based on atomic number priorities. Diastereomers are stereoisomers that are not mirror images, while meso compounds have chiral centers but are achiral due to an internal plane of symmetry.
Chapter 06 an overview of organic reactions
This document discusses organic reaction mechanisms. It explains that reactions occur through a series of steps, and may involve intermediates that are neither the starting reactants nor final products. Polar reactions involve the combination of electrophiles and nucleophiles, while radical reactions involve the formation and reaction of free radicals. The mechanism of HBr addition to ethylene is used as an example, involving the carbocation intermediate. Reaction steps and intermediates are illustrated using energy diagrams, and factors like bond energies and transition states are also discussed.
Chapter 05 stereochemistry at tetrahedral centers
1) The document discusses stereochemistry at tetrahedral carbon centers, including enantiomers, chirality, and how organic molecules can have different mirror image forms.
2) Key concepts covered are how stereochemistry arises from substitution patterns on sp3 hybridized carbon atoms, and how molecules without a plane of symmetry can exist as non-superimposable mirror images called enantiomers.
3) Methods for determining and describing stereochemistry such as sequence rules for assigning R/S configuration at chiral centers and how this relates to optical activity are summarized.
Chapter 05 stereochemistry at tetrahedral centers
Stereochemistry describes 3D properties of molecules that are not identical to their mirror images. These include enantiomers, which are non-superimposable mirror images of each other. Organic molecules containing tetrahedral carbons can have distinct enantiomers. Chiral molecules rotate plane-polarized light and are said to be optically active. Pasteur first discovered distinct crystalline forms of tartaric acid salts that were non-superimposable mirror images. The R/S system assigns configurations at chiral centers based on atomic priorities and spatial orientations. Molecules with multiple chiral centers can also have diastereomers that are not mirror images.
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Section 5.6 logarithmic and exponential equations
- 1. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Section 5.6 Logarithmic and Exponential Equations
- 2. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
- 3. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Recall:
- 4. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. 3 3Solve: log 4 2log x= Reminder: The domain of logarithmic functions is positive numbers only so check for extraneous solutions. 2 3 3log 4 log x= 2 4 x= 2 or 2x x= = −
- 5. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. ( ) ( )2 2Solve: log 2 log 1 1x x+ + − = ( ) ( )2log 2 1 1x x+ − = ( ) ( )2 2 1x x= + − 2 2 2x x= − − + 2 0x x+ = ( )1 0x x + = 0 or 1x x= = − Both solutions are in the domain of the log functions.
- 6. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
- 7. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Recall:
- 8. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Solve: 3 7x = ln3 ln 7x = ln3 ln 7x = ln 7 ln3 x = 1.771≈
- 9. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Solve: 5 2 3x × = 3 2 5 x = 3 ln 2 ln 5 x = ÷ 3 ln 2 ln 5 x = ÷ 3 ln 5 0.737 ln 2 x ÷ = ≈ −
- 10. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. 1 2 3 Solve: 2 5x x− + = 1 2 3 ln 2 ln5x x− + = ( ) ( )1 ln 2 2 3 ln5x x− = + ln 2 ln 2 2 ln5 3ln5x x− = + ln 2 2 ln5 3ln5 ln 2x x− = + ( )ln 2 2ln5 3ln5 ln 2x − = + 3ln5 ln 2 2.186 ln 2 2ln5 x + = ≈ − −
- 11. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Solve: 9 3 6 0x x − − = ( ) 2 Note that 9 = 3 so the equation is quadratic in form.x x 2 3 and = 9x x u u= 2 6 0u u− − = ( ) ( )3 2 0u u− + = 3 or 2u u= = − 3 3 or 3 2x x = = − 1x = 3 0 for allx x>