Quantitative Aptitude - SPLessons

Logarithm Problems

Chapter 24

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Logarithm Problems

shape Introduction

What is a Logarithm? Logarithm is the inverse function of exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. In the simplest case the logarithms denote: How many of one number do we multiply to get another number?. Logarithm Problems chapter discusses the properties of logarithms, common logarithms, its characteristics, & mantissa.


Expression:Logarithm, is, the exponent or power to which a base must be raised to yield a given number. Mathematically: x is the logarithm of n to the base b if \( b^x \) = \( n \) and is represented as \( x \) = \( log_{b}n \)


shape Methods

1. Logarithms: If \( a \) is a positive real number, other than 1 and \( a^m \) = \( x \), then we write: \( m \) = \( log_{a}x \) and say that the value of \( log x \) to the base \( a \) is \( m \).

Example: \( (10)^3 \)= 1000 ⇒ \( log_{10}1000 \) = 3


2.Common logarithms: Logarithms to the base 10 are known as common logarithms.


3.Characteristic:
The integral part of the logarithm of a number is called its characteristic. For example: \( log_{10}15 \) = 1.176 = 1+0.176. So the integer part is 1 and so the characteristic is 1. The characteristic of the logarithm of a number is an integer, negative or positive. For example: \( log_{10}0.5 \) = -0.301 = -1+0.699. Here the characteristic is 1.

(i) When the number is greater than 1.
Here, the characteristic is one less than the number of digits in the left of the decimal point in the given number.


Examples: \( log_{10}15 \) = 1.176 = 1+0.176. Number of digits in 15.0 to the left of decimal is 2. So characteristic is 2-1=1.


\( log_{10}183.5 \) = 2.2636 = 2+0.2636.  Number of digits in 183.5 to the left of decimal is 3. So characteristic is 3-1=2.


(ii) When the number is positive and less than 1.
Here, the characteristic is one more than the number of zeros between the decimal point and the first significant digit of the number and it is negative. The negative number can also be written as: instead of -1, it is written as \(\bar{1}\).


Examples\( log_{10}0.6 \) = -0.2218= -1+0.7782. Number of zeroes in 0.5 to the right of decimal is 1. So characteristic is 0+1=1 but with a negative sign. i.e. characteristic is -1.


\( log_{10}0.008\) = -2.09691001301 = -3+0.9031.  Number of zeroes in 0.008 to the left of decimal is 2. So characteristic is 2+1=3, but with a negative sign. i.e. characteristic is -3.


4.Mantissa: Decimal part of the algorithm of a number is known as its mantissa. Mantissa can never be negative. For example: \( log_{10}15 \) = 1.176 = 1+0.176. The decimal part is 0.176. So, the Mantissa is 0.176. There is also a log table to look for the mantissa.

  1. Product: \(log_{b}(a . c)\) = \(log_{b}a\) + \(log_{b}c\)

  2. Quotient: \(log_{b}\frac{a}{b}\) = \(log_{b}a\) – \(log_{b}c\)

  3. Power: \(log_{b}a^{n}\) = \(n . log_{b}a\)

  4. \(log_{b}1 = 0\)

  5. \(log_{b}b = 1\)

  6. Inverse 1: \(log_{b}b^{n}\) = \(n\)

  7. Inverse 2: \(b{log_{b}n}\) = \(n, n > 0\)

  8. One-to-One: \(log_{b}a\) = \(log_{b}c\) if and only if a = c

  9. Change of Base: \(log_{b}a\) = \(\frac{log_{c}a}{log_{c}b}\) = \(\frac{log a}{log b}\) = \(\frac{In a}{In b}\)

Example 1
Use the properties of logarithms to rewrite expression as a single logarithm:


\(2log_{b}x\) + \(\frac{1}{2}log_{b}(x + 4)\)


Solution:

    \(2log_{b}x\) + \(\frac{1}{2}log_{b}(x + 4)\)

    \(log_{b}x^{2}\) + \(log_{b}(x + 4)^{\frac{1}{2}}\) [Using Power Property]

    \(log_{b}[x^{2}(x + 4)^{\frac{1}{2}}]\) [Using Product Property]


Example 2
Use the properties of logarithms to rewrite expression as a single logarithm:


\(4log_{b}(x + 2) – 3log_{b}(x – 5)\)


Solution:

    \(4log_{b}(x + 2) – 3log_{b}(x – 5)\)

    = \(log_{b}(x + 2)^{4} – log_{b}(x – 5)^{3}\) [Using Power Property]

    = \(log_{b}\frac{(x + 2)^{4}}{(x – 5)^{3}}\) [Using Quotient Property]


Example 3
Use the properties of logarithms to express the following logarithm in terms of logarithm of \(x\), \(y\) and \(z\).


\(log_{b}(xy^{2})\)


Solution:

    \(log_{b}(xy^{2})\) = \(log_{b}x\) + \(log_{b}y^{2}\) [Using Product Property]

    = \(log_{b}x\) + \(2log_{b}y\) [Using Power Property]


Example 4
Use the properties of logarithms to express the following logarithm in terms of logarithm of \(x\), \(y\) and \(z\).


\(log_{b}\frac{x^{2}\sqrt{y}}{z^{5}}\)


Solution:

    \(log_{b}\frac{x^{2}\sqrt{y}}{z^{5}}\)

    = \(log_{b}(x^{2}\sqrt{y}) \ – log_{b}z^{5}\) [Using Quotient Property]

    = \(log_{b}x^{2}\) + \(log_{b}\sqrt{y}\) – \(log_{b}z^{5}\) [Using Product Property]

    = \(2log_{b}x\) + \(\frac{1}{2}log_{b}y\) – \(5log_{b}z\) [Using Power Property]


shape Formulae

Properties of logarithms:

  1. Product: \(log_{b}(a . c)\) = \(log_{b}a\) + \(log_{b}c\)

  2. Quotient: \(log_{b}\frac{a}{b}\) = \(log_{b}a\) – \(log_{b}c\)

  3. Power: \(log_{b}a^{n}\) = \(n . log_{b}a\)

  4. \(log_{b}1 = 0\)

  5. \(log_{b}b = 1\)

  6. Inverse 1: \(log_{b}b^{n}\) = \(n\)

  7. Inverse 2: \(b{log_{b}n}\) = \(n, n > 0\)

  8. One-to-One: \(log_{b}a\) = \(log_{b}c\) if and only if a = c

  9. Change of Base: \(log_{b}a\) = \(\frac{log_{c}a}{log_{c}b}\) = \(\frac{log a}{log b}\) = \(\frac{In a}{In b}\)


shape Examples

1. Calculate the value of: \( log_{3}27 \)?

Solution:

    Given that,

    \( log_{3}27 \)

    Let \( log_{3}27 \) = n

    Then, \(3^n\) = 27

    ⇒ \(3^n\) = \(3^3\)

    ⇒ n = 3

    Therefore, \( log_{3}27 \) = 3


2. If \( log_{\sqrt{8}}x \) = 3 \(\frac{1}{3}\), find the value of \(x\)?

Solution:

    Given that,

    \( log_{\sqrt{8}}x \) = 3 \(\frac{1}{3}\)

    ⇒ \( log_{\sqrt{8}}x \) = \(\frac{10}{3}\)

    ⇒ \( x \) = \((\sqrt{8})^{\frac{10}{3}}\)

    ⇒ \(2^{\frac{3}{2} * \frac{10}{3}}\)

    ⇒ \(2^{5}\)

    ⇒ 32

    Therefore, the value of \(x\) =32


3. Simplify: \(log \frac{75}{16}\) – 2\(log \frac{5}{9}\) + \(log \frac{32}{243}\)?

Solution:

    Given that,

    \(log \frac{75}{16}\) – 2\(log \frac{5}{9}\) + \(log \frac{32}{243}\)

    ⇒ \(log \frac{75}{16}\) – \(log (\frac{5}{9})^2\) + \(log \frac{32}{243}\)

    ⇒ \(log \frac{75}{16}\) – \(log (\frac{32}{243})\) + \(log \frac{32}{243}\)

    ⇒ log 2


4. If \( log_{10}2 \) = 0.30103, find the value of \( log_{10}50 \)?

Solution:

    Given that,

    \( log_{10}2 \) = 0.30103

    Then, \( log_{10}50 \) = \( log_{10}\frac{100}{2} \) = \( log_{10}100 \) – \( log_{10}2 \)= 2 – 0.30103 = 1.69897

    Therefore, \( log_{10}50 \) = 1.69897


5. Simplify : \(\frac{1}{log_{xy}xyz} + \frac{1}{log_{yz}xyz} + \frac{1}{log_{zx}xyz}\)?

Solution:

    Given that,

    \(\frac{1}{log_{xy}xyz} + \frac{1}{log_{yz}xyz} + \frac{1}{log_{zx}xyz}\)

    ⇒ \(log_{xyz}xy + log_{xyz}yz + log_{xyz}xz\)

    ⇒ \(log_{xyz}(xy * yz * zx)\)

    ⇒ \(log_{xyz}(xyz)^2\)

    ⇒ 2 \(log_{xyz}(xyz)\)

    ⇒ 2

    Therefore, \(\frac{1}{log_{xy}xyz} + \frac{1}{log_{yz}xyz} + \frac{1}{log_{zx}xyz}\) = 2


shape Cheat Sheet

Logarithms – Quick Review for Exams


Definition of Logarithms

y = \(\log_{a}\) x if and only if x = \(a^y\), a \(>\) 0, a \(\neq\) 1.

Properties of Logarithms:


1. \(\log_{a}\) 1 = 0


2. \(\log_{a}\) a = 1


3. \(\log_{a}\) 0 = \(\begin{cases}
– \infty\;\;if\;\;a > 1\\
+ \infty\;\;if\;\;a < 1
\end{cases}\)


4. \(\log_{a}\) (xy) = \(\log_{a}\) x + \(\log_{a}\) y


5. \(\log_{a}\;\frac{x}{y}\) = \(\log_{a}\) x – \(\log_{a}\) y


6. \(\log_{a}\;\sqrt[n]{x}\) = \(\frac{1}{n}\) \(\log_{a}\) x


7. \(\log_{a}\;x\) = \(\frac{\log_{c}\;x}{\log_{c}\;a}\) = \(\log_{c}\;x\) \(\cdot\) \(\log_{a}\;c\), c \(>\) 0, c \(\neq\) 1


8. \(\log_{a}\;c\) = \(\frac{1}{\log_{c}\;a}\)


9. x = \(a^{\log_{a}\;x}\)


10. Logarithm to Base 10
      \(\log_{10}\;x\) = log x


11. Natural Logarithm
      \(\log_{e}\;x\) = ln x,
      where e = \(\displaystyle{\lim_{x \to \infty}}(1 + \frac{1}{k})^k\) = 2.718281828…


12. log x = \(\frac{1}{ln 10}\)ln x = 0.434294 ln x


12. ln x = \(\frac{1}{log\;e}\)log x = 2.302585 log x