Logarithm Concepts Properties and Questions for CAT Exam

Logarithm and its properties are very vital concepts for solving many questions from algebra which appear in CAT and other MBA entrance exams. An aspirant who is preparing for CAT must be thorough with the basic properties of logarithm and its working while solving questions.  Let us understand the concepts of logarithm in detail.

Definition of a logarithm: If x>0 and b is a constant \(\left( {b \ne 1} \right)\), then \(y = {\log _b}x\) if and only if \(x = {b^y}\). In the equation \(y = {\log _b}x\), y is referred to as the logarithm, b is the base, and x is the argument.

Properties of logarithm

1. \({\log _a}xy = {\log _a}x + {\log _a}y\)

Proof:

Suppose x=aⁿ and y=a^m

Then the equivalent logarithmic forms are \({\log _a}x = n\) and \({\log _a}y = m\)

We know that \(xy = {a^n} \times {a^m} = {a^{n + m}}\)

Now the logarithmic form of the statement \(xy = {a^{n + m}}\)is \({\log _a}xy = n + m\)substituting the value of n and m we get \({\log _a}xy = {\log _a}x + {\log _a}y\)

2. \({\log _a}\left( {\frac{x}{y}} \right) = {\log _a}x – {\log _a}y\)
3. \({\log _a}{x^m} = m{\log _a}x\)
4. \({\log _{{a^n}}}{x^m} = \frac{m}{n}{\log _a}x\)
5. \({\log _a}x = \frac{1}{{{{\log }_x}a}}\)
6. \({\log _a}a = 1\)
7. \({\log _a}1 = 0\)
8. \({a^{{{\log }_a}m}} = m\)

Example: If \({\log _7}2 = m,\)then \({\log _{49}}28\)is:

Solution:

\({\log _{49}}28 = \frac{{\log 28}}{{\log 49}} = \frac{{\log 7 + \log 4}}{{2\log 7}}\)

=\(\frac{{\log 7}}{{2\log 7}} + \frac{{\log 4}}{{2\log 7}} = \frac{1}{2} + \frac{1}{2}{\log _7}4\)

= \(\frac{1}{2} + \frac{1}{2}.2{\log _7}2\)

=\(\frac{1}{2} + {\log _7}2 = \frac{1}{2} + m\)\( = \frac{{1 + 2m}}{2}\)

Example: If \({\log _e}\left( {\frac{{a + b}}{2}} \right) = \frac{1}{2}({\log _e}a + {\log _e}b)\), then relation between a and b will be

Solution:

\({\log _e}\left( {\frac{{a + b}}{2}} \right) = \frac{1}{2}({\log _e}a + {\log _e}b)\)

\( = \frac{1}{2}{\log _e}(ab) = {\log _e}\sqrt {ab} \)

\( \Rightarrow \frac{{a + b}}{2} = \sqrt {ab} \)

Squaring both side

\(\begin{array}{*{20}{l}}{{{\left( {\frac{{a + b}}{2}} \right)}^2} = ab}\\{ \Rightarrow {{(a + b)}^2} = 4ab}\\{ \Rightarrow {a^2} + 2ab + {b^2} = 4ab}\\{ \Rightarrow {a^2} – 2ab + {b^2} = 0}\\{ \Rightarrow {{(a – b)}^2} = 0}\\{ \Rightarrow a = b}\end{array}\)

Finding number of digits in number of the form a^x.

Step 1: Find the logarithm of a^x in base 10.

Step 2: Take only the integral part of the result and add 1.

Example: If \({\log _{10}}3 = 0.477\), the number of digits in \({3^{40}}\)is

Solution:

Let \(y = {3^{40}}\)

Taking log both sides we get

\(\begin{array}{*{20}{l}}{{{\log }_{10}}y = 40{{\log }_{10}}3}\\{ \Rightarrow {{\log }_{10}}y = 40 \times 0.477 = 19.08}\end{array}\)

The integral part is 19, adding 1 to it we get 19+1 = 20. Hence the number has 20 digits.

Logarithm Inequality:

1. If a>1, p>1 \( \Rightarrow {\log _a}p > 0\)
2. If \(0 < a < 1,p > 1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\log _a}p < 0\)
3. If \(a > 1,{\mkern 1mu} {\mkern 1mu} 0 < p < 1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\log _a}p < 0\)
4. If \(p > a > 1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\log _a}p > 1\)
5. If \(a > p > 1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0 < {\log _a}p < 1\)
6. If \(0 < a < p < 1 \Rightarrow {\log _a}p < 1\)
7. If \(0 < p < a < 1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\log _a}p > 1\)
8. If \({\log _m}a > b{\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left\{ {\begin{array}{*{20}{l}}{a > {m^b},{\mkern 1mu} {\mkern 1mu} {\rm{if}}{\mkern 1mu} {\mkern 1mu} m > 1}\\{a < {m^b},{\mkern 1mu} {\mkern 1mu} {\rm{if}}{\mkern 1mu} {\mkern 1mu} 0 < m < 1}\end{array}} \right.\)
9. \({\log _m}a < b{\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left\{ {\begin{array}{*{20}{l}}{a < {m^b},{\mkern 1mu} {\mkern 1mu} {\rm{if}}{\mkern 1mu} {\mkern 1mu} m > 1}\\{a > {m^b},{\mkern 1mu} {\mkern 1mu} {\rm{if}}{\mkern 1mu} {\mkern 1mu} 0 < m < 1}\end{array}} \right.\)
10. \({\log _p}a > {\log _p}b\) \( \Rightarrow \) \(a \ge b\) if base p is positive and >1 or \(a \le b\)if base p is positive and < 1 e., \(0 < p < 1\)

In other words, if base is greater than 1 then inequality remains same and if base is positive but less than 1 then the sign of inequality is reversed.

Example: If \(x = {\log _3}5,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} y = {\log _{17}}25,\)which one of the following is correct

1. \(x < y\)
2. \(x = y\)
3. \(x > y\)
4. None of these

Solution:

\(y = {\log _{17}}25 = 2{\log _{17}}5\)

\ \(\frac{1}{y} = \frac{1}{2}{\log _5}17\)

\(\frac{1}{x} = {\log _5}3 = \frac{1}{2}{\log _5}9\)

Clearly \(\frac{1}{y} > \frac{1}{x}\) , \$x > y$

Example: If \({\log _{0.3}}(x – 1) < {\log _{0.09}}(x – 1),\)then x lies in the interval

1. \((2,\infty )\)
2. (– 2, –1)
3. (1, 2)
4. None of these

Solution:

\({\log _{0.3}}(x – 1) < {\log _{{{\left( {0.3} \right)}^2}}}(x – 1) = \frac{1}{2}{\log _{0.3}}(x – 1)\)

Or \(\frac{1}{2}{\log _{0.3}}(x – 1) < 0\)

Or \({\log _{0.3}}(x – 1) < 0 = \log 1\) or \((x – 1) > 1\) or \(x > 2\)

As base is less than 1, therefore the inequality is reversed, now x>2 \( \Rightarrow \)x lies in \((2,\infty )\).