ICSE Class 9 Maths Chapter 09 Logarithms

Logarithms

Points To Remember

1. If \(a^b = x\) then \(\log_a x = b\) and it is read as b is the logarithm of x to the base a. Logarithms of only positive real numbers are defined.

Note. (i) \(a^b\) is called the exponential form and \(\log_a x\) is called the logarithmic form.

(ii) \(\log_a 1 = 0\) as \(a^0 = 1\).

(iii) \(\log_a a = 1\) as \(a^1 = a\).

2. Common Logarithms. The logarithms to the base 10 is called the common logarithms and is denoted as \(\log_{10} a\) or simply \(\log a\).

Note. (i) \(\log 10 = 1\) and (ii) \(\log 100 = 2\)

(iii) \(\log 1000 = 3\) and (iv) \(\log 10000 = 4\) etc.

(v) \(\log \frac{1}{10}\) or \(\log 10^{-1}\) or \(\log 0 \cdot 1 = -1\) as \(10^{-1} = \frac{1}{10}\)

(vi) \(\log \frac{1}{100}\) or \(\log 0 \cdot 001\) or \(\log (10)^{-2} = -2\) etc.

Laws Of Logarithms

3. Laws of logarithms:

(i) \(\log_a mn = \log_a m + \log_a n\) (Product law)

(ii) \(\log_a \left(\frac{m}{n}\right) = \log_a m - \log_a n\) (Quotient law)

(iii) \(\log_a (m)^n = n \log_a m\) (Power law)

(iv) \(\log_a a = 1\) and (v) \(\log_a 1 = 0\).

Characteristic And Mantissa Of A Logarithm

4. Characteristic and Mantissa of a logarithm:

A logarithm of a number has two parts:

(i) The integral part which is called characteristic and

(ii) The decimal part which is called mantissa.

Note. (i) The mantissa is always taken as positive while the characteristic may be positive or negative.

(ii) When the characteristic is negative then we denote it by putting a bar on the digit as - 2 is written as \(\overline{2}\)

(iii) 3.5241 means (-3 + .5241)

How To Find The Characteristic And Mantissa Of A Logarithm

5. How to find the characteristic and Mantissa of a logarithm:

Rule 1. The characteristic of logarithm of a number greater than or equal to 1 is one less than the number of digits to the left of the decimal point in the number.

2. The characteristic of logarithm of a number less than 1, is a negative number whose numerical value is one more than the number of zeros immediately following the decimal part.

3. We find mantissa from the log tables.

4. The position of decimal point in a number is immaterial for finding mantissa.

Antilog

6. Antilog: If \(\log m = n\), then antilog \(n = m\).

How To Find Antilog Of A Number

7. How to find antilog of a number.

(i) We use the decimal part for finding the antilog from the antilog table.

(ii) After finding the corresponding number from the antilog table, we insert the decimal point as under:

(a) If characteristic is n, decimal point is put after (n + 1)th digit

If characteristic is \(\overline{n}\) i.e. - n, then decimal point is put in such a way that first significant figure is at the nth place.

Teacher's Note

Logarithms are used in engineering and science to simplify calculations involving very large or very small numbers, making complex computations manageable by hand or with basic calculators.

Exercise 9 (A)

Q. 1. Convert each of the following to logarithmic form:

(i) \(5^2 = 25\)

(ii) \(3^{-3} = \frac{1}{27}\)

(iii) \((64)^{\frac{1}{3}} = 4\)

(iv) \(6^0 = 1\)

(v) \(10^{-2} = 0.01\)

(vi) \(4^{-1} = \frac{1}{4}\)

Q. 2. Convert each of the following to exponential form:

(i) \(\log_3 81 = 4\)

(ii) \(\log_8 4 = \frac{2}{3}\)

(iii) \(\log_2 \frac{1}{8} = -3\)

(iv) \(\log_{10} (0.01) = -2\)

(v) \(\log_5 \left(\frac{1}{5}\right) = -1\)

(vi) \(\log_a 1 = 0\)

Sol. (i) \(\log_3 81 = 4\) and \(\log_4 \left(\frac{1}{4}\right) = -1\) Ans.

(ii) \(5^2 = 25\) \(\therefore \log_5 25 = 2\)

(iii) \(3^{-3} = \frac{1}{27}\) \(\therefore \log_3 \left(\frac{1}{27}\right) = -3\)

(iv) \(6^0 = 1\) \(\therefore \log_6 1 = 0\)

(v) \(10^{-2} = 0.01\) \(\therefore \log_{10} (0.01) = -2\)

(vi) \(4^{-1} = \frac{1}{4}\) \(\therefore\)

Q. 3. By converting to exponential form, find the value of each of the following:

(i) \(\log_2 64\)

(ii) \(\log_8 32\)

(iii) \(\log_3 \frac{1}{9}\)

(iv) \(\log_{0.5} (16)\)

(v) \(\log_2 (0.125)\)

(vi) \(\log_7 7\)

Sol. (i) Let \(\log_2 64 = x\), then \(2^x = 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\)

\(\Rightarrow 2^x = 2^6\)

\(\therefore x = 6\)

Hence \(\log_2 64 = 6\) Ans.

(ii) Let \(\log_8 32 = x\), then \(8^x = 32 \Rightarrow 2^{3x} = 2^5\)

\(\Rightarrow 2^{3x} = 2^5\)

\(\therefore 3x = 5 \Rightarrow x = \frac{5}{3}\)

Hence \(\log_8 32 = \frac{5}{3}\) Ans.

(iii) Let \(\log_3 \frac{1}{9} = x\), then

\(3^x = \frac{1}{9} = \frac{1}{3^2} = 3^{-2}\)

\(\therefore x = -2\)

Hence \(\log_3 \left(\frac{1}{9}\right) = -2\) Ans.

(iv) Let \(\log_{0.5} (16) = x\), then

\((0.5)^x = 16 \Rightarrow \left(\frac{1}{2}\right)^x = 2 \times 2 \times 2 \times 2\)

\(\Rightarrow 2^{-x} = 2^4\)

\(\therefore -x = 4 \Rightarrow x = -4\)

Hence \(\log_{0.5} (16) = -4\) Ans.

(v) Let \(\log_2 (0.125) = x\), then

\(2^x = 0.125 = \frac{0.125}{1000} = \frac{1}{8} = \frac{1}{2^3}\)

\(\Rightarrow 2^x = 2^{-3} \quad \therefore x = -3\)

Hence \(\log_2 (0.125) = -3\) Ans.

(vi) Let \(\log_7 7 = x\), then \(7^x = 7 = 7^1 \quad \therefore x = 1\)

Hence \(\log_7 7 = 1\) Ans.

Teacher's Note

Converting between exponential and logarithmic forms is like translating between two languages - understanding both forms helps solve real-world problems in banking (compound interest) and population growth studies.

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