Logarithm problems play a major role quantitative aptitude test. There is no competitive exam without the questions from this topic. We have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

**PROPERTIES OF LOGARITHM**

(II) log

(III)log

(IV) log

(V) log

(VI)log

(VII) log

CHANGE OF BASE

When we use the log table, we have to remember that the base of "logarithm" must be "10". Because, log table gives values for logarithm only if its base is "10".

Some times, we may have to find value of a "logarithm" whose base is not "10". In that case, we will use the concept "Change of Base".

Let's look at, how the base of a given logarithm can be changed to 10.

For example, if we want to find the value of log

POINTS TO REMEMBER

(B) Thus log 10 = 1, log1 = 0.

(C) Logarithm using base 10 is called common logarithm and logarithm using base "e" is called natural logarithm [e = 2.33(approximately) called exponential number]

RELATION BETWEEN INDICES AND LOGARITHM

If loglog

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from the logarithm problems. Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve logarithm problems. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare this topic. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students have to
learn few basic operations in this topic and some additional tricks. Already we
are much clear with the four basic operations which we often use in math. They
are addition, subtraction, multiplication and division. Even though we are much
clear with these four basic operations, we have to be knowing some more stuff
to do the problems which are being asked from this topic in competitive exams.
The stuff which I have mentioned above is nothing but the tricks and shortcuts
which need to solve the problems in a very short time.

Short cut is nothing but the easiest way to solve problems related to this topic. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as shortcut.

**Here, we are going to have some logarithm problems. You can check your answer online and see step by step solution.**

64 = 2^{6} = 2^{4+2} = 2^{4}x2^{2} = 2^{4}x[(√2)^{2}]^{2}

64 = = 2^{4}x(√2)^{4} = (2√2)^{4}

So, 64 = (2√2)^{4}

log_{2√2}64 = log_{2√2}(2√2)^{4}

log_{2√2}64 = 4log_{2√2}(2√2)

log_{2√2}64 = 4(1)

log_{2√2}64 = 4

Hence, the value of log 64 to the base 2 √2 is 4

64 = = 2

So, 64 = (2√2)

log

log

log

log

Hence, the value of log 64 to the base 2 √2 is 4

2. If log

1/(x+1) + 1/(y+1) + 1/(z+1) is

x + 1 = log_{a}bc + log_{a}a = log_{a}abc

y + 1 = log_{b}ca + log_{b}c = log_{b}abc

z + 1 = log_{c}ab + log_{c}c = log_{c}abc

1/(x+1) = 1/log_{a}abc = log_{abc}a

1/(y+1) = 1/log_{b}abc = log_{abc}b

1/(z+1) = 1/log_{c}abc = log_{abc}c

1/(x+1)+1/(y+1)+1/(z+1) = log_{abc}a + log_{abc}b + log_{abc}c

1/(x+1)+1/(y+1)+1/(z+1) = log_{abc}abc

1/(x+1)+1/(y+1)+1/(z+1) = 1

y + 1 = log

z + 1 = log

1/(x+1) = 1/log

1/(y+1) = 1/log

1/(z+1) = 1/log

1/(x+1)+1/(y+1)+1/(z+1) = log

1/(x+1)+1/(y+1)+1/(z+1) = log

1/(x+1)+1/(y+1)+1/(z+1) = 1

3. If a = log

1 + abc = 1 + log_{24}12xlog_{36}24xlog_{48}36

1 + abc = 1 + log_{36}12xlog_{48}36

1 + abc = 1 + log_{48}12

1 + abc = log_{48}48 + log_{48}12

1 + abc = log_{48}(48x12)

1 + abc = log_{48}(2x12)^{2}

1 + abc = 2log_{48}24

1 + abc = 2log_{36}24xlog_{48}36

1 + abc = 2bc

1 + abc = 1 + log

1 + abc = 1 + log

1 + abc = log

1 + abc = log

1 + abc = log

1 + abc = 2log

1 + abc = 2log

1 + abc = 2bc

4. The value of log 0.0001 to the base 0.1 is

log_{0.1}0.0001 = log_{0.1}(0.1)^{4}

log_{0.1}0.0001 = 4log_{0.1}0.1

log_{0.1}0.0001 = 4(1)

log_{0.1}0.0001 = 4

log

log

log

5. If 2 log x = 4 log 3, then the value of "x" is

2 log x = 4 log 3

log x = (4 log 3)/2

log x = 2 log 3

log x = log 3^{2}

log x = log 9

Removing "log" on both the sides, we get

x = 9

log x = (4 log 3)/2

log x = 2 log 3

log x = log 3

log x = log 9

Removing "log" on both the sides, we get

x = 9

6. The value of log

log_{√2}64 = log_{√2}(2)^{6}

log_{√2}64 = 6 log_{√2}(2)

log_{√2}64 = 6 log_{√2}(√2)^{2}

log_{√2}64 = 6x2 log_{√2}(√2)

log_{√2}64 = 12x1

log_{√2}64 = 12

log

log

log

log

log

7. The value of log (1/81) to the base 9 is equal to

log_{9}(1/81) = log_{9}1 - log_{9}81

log_{9}(1/81) = 0 - log_{9}(9)^{2}

log_{9}(1/81) = -2 log_{9}(9)

log_{9}(1/81) = -2x1

log_{9}(1/81) = -2

log

log

log

log

8. log 0.0625 to the base 2 is equal to

log_{2}0.0625 = log_{2}(0.5)^{4}

log_{2}0.0625 = 4 log_{2}(0.5)

log_{2}0.0625 = 4 log_{2}(1/2)

log_{2}0.0625 = 4 [log_{2}1 - log_{2}2]

log_{2}0.0625 = 4 [0 - 1]

log_{2}0.0625 = -4

log

log

log

log

log

9. Given log 2 = 0.3010 and log 3 = 0.4771, the value of log 6 is

log 6 = log (2x3)

log 6 = log 2 + log 3

log 6 = 0.3010 + 0.4771

log 6 = 0.7781

log 6 = log 2 + log 3

log 6 = 0.3010 + 0.4771

log 6 = 0.7781

10. The value of log 0.3 to the base 9 is

log_{9}0.3 = log_{9}(1/3)

log_{9}0.3 = log_{9}1 - log_{9}3

log_{9}0.3 = 0 - log_{9}3

log_{9}0.3 = -1/log_{3}9

log_{9}0.3 = -1/log_{3}(3)^{2}

log_{9}0.3 = -1/2log_{3}(3)

log_{9}0.3 = -1/2(1)

log_{9}0.3 = -1/2

log

log

log

log

log

log

log

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