Logarithms Notes for Competitive Exams Preparation

Logarithms Notes for Competitive Exams Preparation

Logarithms:

In the Equation a^x = N, we are expressing N in terms of a and x. The same equation can be re-written as a = N ^1/x . Here we are expressing "a" in terms of N and x. But, among a, x and N, by normal algebraic methods known to us, we cannot express x in terms of the other two parameters a and N. This is where logarithms come into the picture. When a^x = N, then we say x= logarithm of N to the base a, and write it as x= log_aN. The definition of logarithm is given as "the logarithm of any number to a given base is the index of the power to which the base must be raised on order in order to equal the given number".

Thus, if a^x = N then x = log _a N(log N to the base a). This is the basic definition of a logarithm.

This basic definition of logarithm is very useful in solving a number of problems on logarithms.

Example of a logarithm : 216 = 6³ can be expressed as log ₆ 216 = 3.

Since logarithm of a number is a value, it will have an "integral" part and a "decimal" part. The integral part of the logarithm of a number is called the Characteristic and the decimal part of the logarithm is called the Mantissa.

Logarithms are defined only for positive number. There are no logarithms defined for ZERO or Negative Numbers.

Logarithms can be expressed to any base. Logarithms from one base can be converted to logarithms to any other base. (One of the formulae given below will help do this conversion). However, there are two types of logarithms that are commonly used.

i) Natural Logarithms or Napierian Logarithms:
These are logarithms expressed to the base of a number called "e"

ii) Common Logarithms:
These are logarithms expressed to the base 10. For most of the problems under Logarithms, it is common logarithms that we deal with. In examinations also, if logarithms are given without mentioning any base, it can normally be taken to be logarithms to the base 10.

 Given below are some Important Basic Rules/Formulae in logarithms:

1. log _a a=1 (logarithm of any number to the same base is 1)
2. log _a1=0 (log of 1 to any base is 0)
3. log _a( mn ) = log_a m +log_a n
4. log _a(m/n) = log _am-log_an
5. log _am ^p = p log _a m
6. log _a ^q m ^p = p/q log _a m
7. log _a m = log _bm / log _ba
8. log _a b = 1 / log _ba
9. a ^log _a N = N

Worked out Examples:

1) If log (x-3)+log(x+2)= log 6 , then find the value of x.

Sol:-
           log(x-3)(x+2) = log 6

(x-3)(x+2) = 6
x²-x-12=0
(x-4) (x-3) = 0
x = 4 or x = -3
If x = -3, log(x+2) does not exist, as logarithms are not defined for negative values.

Therefore x = 4.

2) Prove that a=b, if 2 log 2+log a+log b=2 log (a+b).

sol:-
         log (2² x a  x b)=log (a+b)²
4ab = a² + b² +2ab
a² + b² - 2ab = 0
(a-b)² = 0
a-b = 0
Therefore a = b.

3) If log (3x - 5) = log (2x+3) find x.

Sol:- 
         log(3x-5) = log (2x+3)
3x-5 = 2x+3
x= 8.

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Averages Mixtures Allegations Notes for Competitive Exams Preparation

Averages Mixture Allegation Notes

Average:

Average is a very simple but effective way of representing an entire group by a single value.

"Average " of a group is defined as

Average = Sum of all items in the group/Number of items in the group

A student expresses his score in the exams by giving "average" percentage in all paper together rather than scores in individual papers.

Similarly, if there are 60 students in a class, instead of talking of the height of each individual student, we can talk of "average" height of the class.

Note:- Average of any number of quantities will always be higher than the lowest value and lower than the higher value of items whose average is being taken.

Mixture- Allegation:

Mixing of two or more types of qualities of things produces a mixture. When two items of different qualities are thus mixed, the quality of the resultant mixture lies in between the qualities of the original constituent items, i.e., it will be higher than the lowest quality and lower than the highest quality of the items being mixed.

Similarly, if two types of a product of different prices per unit are mixed, the unit price of the resultant mixture will lie between the prices of the two types that form the mixture.

The following Rule of Allegation connecting quantities and mixture will be useful in problems on mixtures and weighted averages.

Quantity of cheaper ₌ Rate of dearer -Average Rate    
Quantity of dearer       Average Rate - Rate of Cheaper

Worked out Examples:

1) The daily temperature in a week from monday to sunday is 42°C, 43° C, 42°C, 41°C, 40°C, 44°C, 42°C respectively. What is the daily average temperature ?

Sol:-
        Average Temperature ₌ Sum of the temperatures
                                                        Number of Days

 =  42+43+42+41+40+44+42 
               7
= 42°C

2) The cost of none quality of rice is Rs. 12 per kg and that of other quality is Rs. 16 per kg. If six kilograms of the first quality and four kilograms of the second quality are mixed, what is the average cost of the mixture ?

Sol:-
         The cost of 4 kg of first quality of rice = 6x12 = Rs.72
         The cost of second quality of rice = 4x16= Rs.64
         Total cost of 10 kgs = 72+64 = Rs.136
         Average cost of the mixture = 136/10 = Rs.13.60

Note:- This average, as detailed in the above example is called " the Weighted Average". The price of rice is "weighted" by the quantity purchased to get "Weighted Average Price " of rice.

3) How much quantity of tea costing Rs.120 a kg be mixed with 6 kgs of tea costing Rs. 180 a kg such that the mixture costs Rs. 160 a kg ?

sol:- 
        Applying the rule of allegation

Quantity of Rs.120 a kg = 180-160
Quantity of Rs.180 a kg     160-120

= 20/40= 1/2

Therefore the quantity of tea costing Rs.120 a kg that is mixed = (1/2)6 = 3 kg


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Ratio-Proportion-Variation Notes for Competitive Exams Preparation

Ratio-Proportion-Variation Notes for Competitive Exams Preparation

Ratio:

Ratio is the relation which one quantity bears to another of the same kind, the comparison being made by considering what multiple, part or parts, one quantity is of the other. The ration of two quantities "a" and "b" is represented as a : b and read as "a is to b". "a" is called antecedent, "b" is the consequent. Since the ratio expresses the number of times one quantity contains the other, its's an abstract quantity.



A ratio a : b can also be expressed as a/b. So if two items are in the ratio 2 : 3, we can say that their ratio is 2/3. If two terms are in the ratio 2, it means that they are in the ratio of 2/1, ie., 2 :1.

" A ratio is said to be a ratio of greater inequality of less inequality or of equality according as antecedent is greater than, less than or equal to consequent". From this we find that a ratio of greater inequality is diminished and ratio of less inequality is increased by adding same quantity to both terms.

ie., in a : b
If a<b then (a+x) : (b+x)> a: b and a>b then (a+x):(b+x) < a:b

if a/b=c/d=e/f ...., then each of these ratio is equal to a+c+e+..../b+d+f+...

Proportion:

When two ratios are equal the four quantities involved in the two ration are said to be proportional
ie., If a/b= c/d, then a, b, c and d are proportional. This is represented as a:b :: c:d and si read as "a is to b (is) as c is to d".

When a, b, c and d are in proportion, then the items a and d are called the EXTREMES and the items b and c are called the MEANS. We also have the relationship.

Product of the MEANS = Product of the EXTREMES ie., bc= ad

If a:b = c:d then
b:a = d:c----(1)
a: c= b:d
a+b: b= c+d:d....(2) obtained by adding 1 to both sides of the given relationship
a-b:b=c-d:d.....(3) obtained by subtracting 1 from both sides of the given relationship
a+b:a-b=c+d:c-d .....(4) obtained by dividing relationship (2) above by (3).

Relationship
(1) above is called INVERTENDO
(2) is called COMPONENDO
(3) is called DIVIDENDO and
(4) is called COMPONENDO- DIVIDENDO.

The last relationship, ie., COMPONENDO-DIVIDENDO is very helpful in simplifying problems.

Whenever we know a/b=c/d, then we can write (a+b)/(a-b)=(c+d)/(c-d) by this rule. The converse of this is also true whenever we know that (a+b)/(a-b)= (c+d)/(c-d), then we can conclude that a/b=c/d.

If three quantities a, b and c are such that a:b :: b:c , then we say that they are in CONTINUED PROPORTION. We also get b square= ac

Variation:

Two quantities A and B may be such that as one quantities changes in value, the other quantity also changes in value depending on the change in the value of the first quantity.

Direct Variation:

One quantity A is said to vary directly as another quantity B if the two quantities depend upon each other in such a manner that if B is increased in a certain ration, A is increased in the same ratio and if B is decreased, A is decreased in the same ratio.

This is denoted as A alpha B then A= kB, where k is a constant. It is called a constant of proportionality.

Inverse Variation:

A quantity A is said to vary inversely as another quantity B if the two quantities depend upon each other in such a manner that if B is increased in a certain ratio, A is decreased in the same ratio and if B is decreased then A is increased in the same ratio. It is the same as saying that A varies directly with 1/B. It is denoted as if A alpha 1/B i.e., A=k/B, where k is a constant of proportionality.

Joint Variation:

If there are three quantities A, B, and C such that A varies with B when C is constant and varies with C when B is constant, then A is said to vary  jointly with B and C when both B and C are varying.

then A alpha BC or A=kBC, where k is the constant of proportionality.

Worked Out Examples:

1) Two numbers are in the ratio 9: 11. If 5 be added to each, then the numbers will be in the ratio 5: 6. Find the numbers.

Sol:- x/y= 9/11
Let x= 9k, y= 11k
9k+5/11 k+5=5/6
54k+30=55 k+25
k= 5
x=9(5)=45
y=11(5)= 55

2) The volume of cylinder varies jointly as its height and the area of its base. When the area of the base is 300 square feet and the height is 6 feet the volume is 600 cu.ft. what is the area of the base of a cylinder whose volume is 360 cu.ft. and height is 12 ft.

Sol:-  Let v be the volume, a be the area of base, h be the height.

V=m a h, where m is the constant.
Therefore, 600=m(300)(6)
1/3=m
therefore V=1/3(a)(h)
Now, 360 = 1/3(a)(12)
a= 90 sq.ft.

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