JKBOSE 9th Class Mathematics Solutions Chapter 4 Logarithms

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Jammu & Kashmir State Board JKBOSE 9th Class Mathematics Solutions

J&K class 9th Mathematics Logarithms Textbook Questions and Answers

Rational Numbers As Exponents :

(i) If m is a positive integer and x and y are rational numbers such that x = y, then y^1/m = x.

y^1/m is called the mth root of y and is written as m√y.

Using this meaning we define x^m for a positive rational exponent m. If m= 0 then x⁰ = 1.

(ii) If x = r/s is a non-zero rational number, then {r/s}^-m = {s/r}^m

Rational Numbers As Exponents When Exponent is Any Positive Rational Number

If x is a positive rational number and m = p/q is a positive rational Pexponent then we define x^p/q as:

(i) the qth root of x^p i.e.

TEXT BOOK EXERCISE – 4.1

1. Simplify each of the following removing redical signs nagative indices wherever they occur :

2. Assuming that x, y, z are positive real numbers, simplify each of the following :

Solution.

3. Assuming that x, y, z are positive real numbers and exponents are all rational numbers, show that :

Solution.

Definition of Logarithms

For a positive real number a and a rational number m Let a^m = b where b is a real number. In other words the mth power of base a is b. Another way of stating the same fact is logarithm of b to base a is m. We write it as log_a^b = m “log” being the abbreviation of the word “logarithm”. 1

Note : log₁₀1 = 0 since 10⁰ = 1

TEXT BOOK EXERCISE – 4.2

1. Write the following in the form of logarithms :

(i) 2⁵ = 32 (ii) 10³ = 1000 (iii) 3⁴ = 81 (iv) 5⁴ = 625 (v) 10^-1 = 0.1 (vi) 7² = 49

Solution.— (i) 2⁵= 32 Here a = 2, m = 5 and b = 32 the 5th power of base 2 is 32.

In form of logarithms, the logarithm of 32 to base 2 is 5.

Thus we have log₂32 = 5.

(ii) 10³ = 1000 Here a = 10, m = 3 and b = 1000. The 3rd power of base 10 is 1000.

In form of logarithms; the logarithm of 1000 to the base 10 is 3.

Thus we have log₁₀1000 = 3

(iii) 3⁴ = 81, here a = 3, m = 4 and b = 81, the 4th power of base 3 is 81.

In the form of logarithms; the logarithm of 81 to the base 3 is 4.

Thus we have log₃81 = 4.

(iv) 5⁴ 625, here a = 5, m = 4 and b = 625, the 4th power of the base 5 is 625.

In the form of logarithms; the logarithm of 625 to the base 5 is 4.

Thus we have log₅625 = 4.

(v) 10^-1= 0.1, here a = 10, m = – 1 and b = 0.1, the – 1 power of the base 10 is 0.1.

In the form of logarithms; the logarithm of 0.1 to the base 10 is -1.

Thus we have log₁₀ (0.1) = − 1

(vi) 7² = 49. Here a = 7, m = 2 and b = 49.

The 2nd power of the base 7 is 49.

In the form of logarithms; the logarithm of 49 to the base 7 is 2.

Thus we have log₇ 49 = 2.

2. Express each of the following in exponential form :

(i) log₅25 = 2

(ii) log₃ 243 = 5

(iii) log₁₀1000 = 3

(iv) log₂ 64 = 6

(v) log₄ 64 = 3

Solution.— (i) log₅25 = -2; Here b = 25, a = 5 and m = 2

We write log_a b = m in exponential form, as a^m = b

So we write log₅25 = 2 in exponential form as 5² = 25

(ii) log₃243 = 5; Here b = 243, a = 3 and m = 5

We write log_ab = m in exponential form; as a^m = b

So we write log₃243 = 5 in exponential form as 3⁵ = 243

(iii) log₁₀1000 = 3; Here a = 10, m = 3 and b= 1000 in exponential form we write log_ab = m as a^m

So we write log₁₀1000 = 3 in the exponential form as 10³ = 1000

(iv) log₂64 = 6; Here a = 2, m = 6 and b = 64

We write log_ab = m in exponential form; as a^m = b

So we write log₂64 = 6 in exponential form as 2⁶ = 64.

(v) log₄64 = 3; Here a = 4, m = 3 and b = 64

We write log₄b = m in exponential form as a^m = b

So write log₄64 = 3 in the exponential form as 4³ = 64

Laws of Logarithms

TEXT BOOK EXERCISE – 4.3

In each of the following, assume that the base a = 10 wherever, it has not been indicated :

2. Suppose that

3. We can write

4. We can write

Standard Form of Decimal

Any positive decimal can be wirtten in the form

n = m × 10^P

where p is an integer (positive, zero or negative) and 1 ≤ m < 10.

This is called the standard form of n.

Working Rule saiety

1. Move the decimal point to the left or to the right, as may be necessary, to bring one non-zero digit to the left of decimal point.

2. (a) If we move p places to the left; multiply by 10^P.

(b) If we move p places to the right, multiply by 10^-P.

(d) Write the new decimal obtained by the power of 10 (of step 2) to obtain the standard form of the given decimal.

TEXT BOOK EXERCISE – 4.4

1. Write each of the following in standard form :

(i) 3.12 (ii) 31.23 (iii) 312.3 (iv) 3123 (v) 312300 (vi) 0.3123 (vii) .03123.

Solution.— As we know that any positive decimal n is in its standard form if we express it as n = m × 10^P.

where p is an integer (positive, zero or negative) and 1 ≤ m < 10.

Now;

2. Write the following numbers in decimal form, without powers of 10 as factors :

(i) 5.6 × 10³

(ii) 1.436 × 10-¹

(iii) 2.4 x 10-²

(iv) 9.632 × 10⁵

(v) 1.2056 × 10²

(vi) 1.2056 × 10-².

Solution.—

Characteristic and Mantissa

Consider the standard form of n

n = m × 10^P where 1 ≤ n < 10.

Taking logarithms to the base 10 and using the laws of logarithms.

log n = log (m × 10^P)

= log m + log 10^P

= log m + p log 10

= log m + p.

or log n = p + log m

Here p is an integer and called Characteristic of log n.

As 1 ≤ m < 10, so 0 ≤ log m < 1

log m is called mantissa of log n.

Step to find log n

1. Put n in the standard form, say

n = mx 10^P, 1 ≤ m < 10

2. Read off the characteristic p of log n from this expression (exponent of 10).

3. Look up log m from the tables

For example for log (6.234), we look in row 62 under column 3. We find 7945.

So 7945 + 3 = 7948

Hence log (6.234) = 0.7948.

4. Write log n = p + log m.

TEXT BOOK EXERCISE – 4.5

1. Use logarithm tables to find the logarithms of the following numbers :

(i) 1270

(ii) 12.70

(iii) 431.5

(iv) 11.23

(v) 0.1257

(vi) 0.0012

(vii) 0.00001379.

Finding n when log n is given

If log n = t, we sometime say = antilog t

we write log n as

log n = p + log m

We first take antilog of mantissa; log m in antilog table. Let the number corresponding to log m be k i.e. antilog (log m) = k. Since the characteristic of log n is p. So the standard form of n is given by n = k × 10^P

TEXT BOOK EXERCISE – 4:6

1. Using tables, find the logarithm of each of the following numbers :

(i) 4.8

(ii) 7

(iii) 3.17

(iv) 3.172

(v) .235

(vi) .2354

2. Find log x, if x equals

(i) .0768

(ii) .0025

(iii) .0087

(iv) .00954

(v) .0056

(vi) .0287

3. Find the antilogarithm of each of the following :

(i) 0.752

(ii) .301

(iii) .5428

(iv) 2.752

(v) 1.301

(vi) 2.5428

4. Each of the following numbers is logarithm of some number. Express each in the form p + log in, where p is the characteristic and log m the mantissa and find the number.

(i) 1.2086

(ii) – 1.2084

(iii) – 2.4325

(iv) – 3.6432

(v) 2.5674

(vi) – 0.62.

Use of logarithms in numerical calculation

As we know that logarithm is a rule to shorten lengthy and difficult calculations; in this section we shall use logarithms in numerical calculation.

TEXT BOOK EXERCISE – 4.7

(Use logarithmic tables)

1. Find the value of

(i) 6.45 × 981.4

(ii) 0.0064 × 1.507

2. Evaluate :

(i) 2.632/.0045

(ii) 0.54 ÷ 216.3

3. Find the value of :

(i) (.724)³

(ii)√42.36

4. Find the cube root of 48 correct to two decimal places.

Solution.— Let x be the cube root of 48.

∴ x = 3√48

or x = (48)^1/3

Taking log both sides; we get” :

log x = log (48)^1/3

= 1/3 log 48 [Using log mⁿ = n log m]

= 1/3 × 1.6812

log x = 0.5604

Taking antilog both sides, we get :

x = antilog (0.5604)

x = 3.634

⇒ x = 3.63 correct up to two places of decimal.

5. Write down the logarithm of ⁶⁴ and use it to state the number of digits in the numeral for 2⁶⁴.

Solution.— Let x = 2⁶⁴

Taking log both sides; we get :

log x = log 2⁶⁴

= 64 log 2 [Using log mⁿ = n log m]

= 64 × .3010

log x= 19.264

To find the number of digits in numeral,

we take antilog both sides; we get :

x= antilog (19.264)

= 1.837 × 10¹⁹

or x = 1837 × 10¹⁶

Hence number of digits in 2⁶⁴ = 20

ALTER. To find the number of digits in numeral :

We have log x= 19.264

∴ Characteristic of log 2⁶⁴= 19

∴ Number of digits in 2⁶⁴ = 19 + 1 = 20

6. Using logarithmic tables; evaluate :

7. Find the values of the following :

(a) 3√0.0847

(b) (0.09634)³

(c) 4√.6789

(d) 5√42.7

Taking antilog both sides ; we get :

x = antilog × (0.3260)

= 2.118 × 10⁰

x = 2.118

8. Using logarithms; simplify :

(a) 57.12 × 2.034

(b) 0.8623 × 0.000451

(c) 352.6 × 0.078 × 0.5943

(d) 2456 × 0.000071

(e) 328.4 × 12.65

(f) .3865 ÷ 0.000572

(g) 0.0953 + 3.794

(h) 25 ÷ 0.0683

9. Find the values of the following :

TEXT BOOK EXERCISE – 4.8

1. A savings bank account pays 5% interest (per year) and it compounded every six months. When a boy is 13 years old, ₹ 100 is deposited to his credit in a saving, bank account. How much is due to him when he is 21 years old ?

2. In how many years will an account double itself at 5% interest compounded annually ?

3. A nationalised bank issues “Re-investment Certificates” for a period of 3 years. If ₹ 5000 is invested in these certificates, their maturity value is ₹ 6725. Assuming that the interest is compounded every year, what is the rate of interest ?

Solution.— Let Principal sum; P = ₹ 5000

Amount; A = ₹ 6725 was

Time; n = 3 years

Let rate of interest be r% per annum

Now using the formula of compound interest;

4. The cash price of a new car is ₹ 90,000. The insurance company calculates its price at any subsequent time according to the rule that the price depreciates at the rate of 5% a year during the first two years and at the rate of 10% a year thereafter. What will be the price of the car after

(a) 2 years (b) 5 years (c) 10 years.

5. The cash price of a refrigerator is ₹ 5500. It can also be purchased for ₹ 2000 followed by 11 equal monthly instalments of ₹ 400 each. What is the rate of interest charged under instalment plan?