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Real Numbers Class 10th Mathematics AP Board Solution

Class 10th Mathematics AP Board Solution

Exercise 1.1
Question 1.

Use Euclid’s division algorithm to find the HCF of

900 and 270

Euclid’s Division is a method for finding the HCF (highest common factor) of two given integers. According to Euclid’s Division Algorithm, For any two positive integers, ‘a’ and ‘b’, there exists a unique pair of integers ‘q’ and ‘r’ which satisfy the relation:

a = bq + r , 0 ≤ r ≤ b

Given integers 900 and 270. Clearly 900>270.

By applying division lemma

⇒ 900 = 270×3 + 90

Since remainder 0, applying division lemma on 270 and 90

⇒ 270 = 90×3 + 0

∵ remainder = 0,

∴ the HCF of 900 and 270 is 90.

Question 2.

Use Euclid’s division algorithm to find the HCF of

196 and 38220

Euclid’s Division is a method for finding the HCF (highest common factor) of two given integers. According to Euclid’s Division Algorithm, For any two positive integers, ‘a’ and ‘b’, there exists a unique pair of integers ‘q’ and ‘r’ which satisfy the relation:

a = bq + r , 0 ≤ r ≤ b

Given integers 196 and 38220. Clearly 38220>196.

By applying division lemma

⇒ 38220 = 196×195 + 0

Since remainder 0

∴ the HCF of 196 and 38220 is 195.

Question 3.

Use Euclid’s division algorithm to find the HCF of

1651 and 2032

Euclid’s Division is a method for finding the HCF (highest common factor) of two given integers. According to Euclid’s Division Algorithm, For any two positive integers, ‘a’ and ‘b’, there exists a unique pair of integers ‘q’ and ‘r’ which satisfy the relation:

a = bq + r , 0 ≤ r ≤ b

Given integers 1651 and 2032. Clearly 2032>1651.

By applying division lemma

⇒ 2032 = 1651×1 + 381

Since remainder 0, applying division lemma on 1651 and 381

⇒ 1651 = 381×4 + 127

Since remainder 0, applying division lemma on 381 and 127

⇒ 381 = 127×3 + 0

Since remainder = 0,

∴ the HCF of 1651 and 2032 is 127.

Question 4.

Use Euclid division lemma to show that any positive odd integer is of form 6q + 1, or 6q + 3 or 6q + 5, where q is some integers.

Let a be any odd positive integer and b = 6

Then using Euclid’s algorithm, we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0,1,2,3,4,5because 0≤r<band the value of b is 6

So total form available will be 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5 ,6q + 6 is divisible by 2, so it is an even number.

6q + 1, 6 is divisible by 2 but 1 is not divisible by 2, so it is an odd number.

6q + 2 , 6 is divisible by 2 but 2 is also divisible by 2, so it is an even number.

6q + 3, 6 is divisible by 2 but 3 is not divisible by 2, so it is an odd number.

6q + 4, 6 is divisible by 2 but 4 is also divisible by 2, so it is an even number.

6q + 5, 6 is divisible by 2 but 5 is not divisible by 2, so it is an odd number.

∴ so odd numbers will in form of 6q + 1 or 6q + 3 or 6q + 5

Question 5.

Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1 or 3p + 2.

Let be any positive integer. Then, it is form 3q or, 3q + 1 or, 3q + 2

So, we have the following cases:

Case I When a = 3q

In this case, we have

a2 = (3q)2 = 9q2 = 3q(3q) = 3p, where p = 3q2

Case II When a = 3q + 1

In this case, we have

a2 = (3q + 1)2 = 9q2 + 6q + 1 = 3q(3q + 2) + 1 = 3p + 1,

where p = q(3q + 2)

Case III When a = 3q + 2

In this case, we have

a2 = (3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1

= 3(3q2 + 4q + 1) + 1 = 3p + 1

where p = 3q2 + 42 + 1

Hence, a is the form of 3p or 3p + 1 or 3p + 2

Question 6.

Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Let a be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

We know that according to Euclid's division lemma:
a = bq + r
So, we have the following cases:

Case I When a = 3q

In this case, we have

a3 = (3q)3 = 27q3 = 9(3q3 ) = 9m, where m = 3q3

Case II When a = 3q + 1

In this case, we have

a3 = (3q + 1)3

⇒ 27q3 + 27q2 + 9q + 1

⇒9q(3q2 + 3q + 1) + 1

⇒ a3 = 9m + 1, where m = q(3q2 + 3q + 1)

Case III When a = 3q + 2

In this case, we have

a3 = (3q + 1)3

⇒ 27q3 + 54q2 + 36q + 8

⇒9q(3q2 + 6q + 4) + 8

⇒ a3 = 9m + 8, where m = q(3q2 + 6q + 4)

Hence, a3 is the form of 9m or, 9m + 1 or, 9m + 8

Question 7.

Show that one and only one out of n, n + 2 or n + 4is divisible by 3, where n is any positive integer.

We know that any positive integer is of the form 3q or, 3q + 1 or,

3q + 2 for some integer and one and only one of these possibilities can occur.

So, we have following cases:

Case I When n = 3q

In this case, we have

n = 3q, which is divisible by 3

Now,

⇒ n + 2 = 3q + 2,

⇒ n + 2 leaves remainder 2 when divided by 3

⇒ n + 2 is not divisible by 3

Again, n = 3q

⇒ n + 4 = 3q + 4 = 3(q + 1) + 1

⇒ n + 4leaves remainder 1 when divided by 3

⇒ n + 4 is not divisible by 3

Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3.

Case II When n = 3q + 1

In this case, we have

n = 3q + 1

⇒ n leaves remainder 1 when divided by 3

⇒ n is not divisible by 3

Now, n = 3q + 1

⇒ n + 2 = (3q + 1) + 2 = 3(q + 1),

⇒ n + 2 is divisible by 3

Again, n = 3q + 1

⇒ n + 4 = (3q + 1) + 4 = 3q + 5 = 3(q + 1) + 2

⇒ n + 4 leaves remainder 2 when divided by 3

⇒ n + 4 is not divisible by 3

Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3.

Case III When n = 3q + 2

In this case, we have

n = 3q + 2

⇒ n leaves remainder 2 when divided by 3

⇒ n is not divisible by 3

Now, n = 3q + 2

⇒ n + 2 = 3q + 2 + 2 = 3(q + 1) + 1,

⇒ n + 2 leaves remainder 1 when divided by 3

⇒ n + 2 is not divisible by 3

Again, n = 3q + 2

⇒ n + 4 = 3q + 2 + 4 = 3(q + 2)

⇒ n + 4 is divisible by 3

Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3.

Exercise 1.2
Question 1.

Express each of the following number as a product of its prime factors.

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

I. 140 = 2×2×5×7 = 22×5×7

II. 156 = 2×2×3×13 = 22×3×13

III. 3825 = 3×3×5×5×17 = 32×52×17

IV. 5005 = 5×7×11×13

V. 7429 = 17×19×23

Question 2.

Find the LCM and HCF of the following integers by the prime factorization method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

(iv) 72 and 108

(v) 306 and 657

I. 12,15 and 21

12 = 22×3

15 = 3×5

21 = 3×7

LCM = 22×3×5×7 = 420

HCF = 3

II. 17,23 and 29

17 = 1×17

23 = 1×23

29 = 1×29

LCM = 1×17×23×29 = 11339

HCF = 1

III. 8, 9 and 5

8 = 23

9 = 32

5 = 1×5

LCM = 23×32×5 = 360

HCF = 1

IV. 72 and 108

72 = 23×32

108 = 22×33

LCM = 25×35 = 7776

HCF = 22×32 = 4×9 = 36

V. 306 and 657

306 = 2×32×17

657 = 32×73

LCM = 2×32×17×73 = 22338

HCF = 32 = 9

Question 3.

Check whether 6n can end with the digit 0 for any natural number n.

If any number end with digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as

Prime factorization of 6n = (2×3)n

It can be observed that 5 is not in the prime factorization of 6n.

Hence, for an value of n, 6n will not visible by 5.

∴ 6n cannot end with the digit 0 for any natural number n.

Question 4.

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Numbers are of two types – composite and prime. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

7×11×13 + 13 = 13×(7×11 + 1) = 13×(77 + 1) = 13×78

= 13×13×6

The given expression has 6 and 13 as its factors.

∴ , it is a composite factor.

7×6×5×4×3×2×1 + 5 = 5×(7×6×4×3×2×1 + 1)

= 5×(1008 + 1) = 5×1009

1009 cannot be factorized further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Question 5.

How will you show that (17 × 11 × 2) + (17 × 11 × 5) is a composite number? Explain.

Numbers are of two types – composite and prime. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

(17×11×2) + (17×11×5) = 17{(11×2) + (11×5)}

= 17×{11×{(2) + (5)}} = 17×11×7

The given expression has 17, 11and 7 as its factors.

∴ it is a composite factor.

Question 6.

What is the last digit of 6100.

This is related to concept of numbers in the unit digits place of the powers of natural number. The power of 6 any index repetition 6 i.e. (6)n the last digit is 6 only.

Example:

i: 61 = 6

ii: 62 = 36

iii: 63 = 216

The last digit in the expansion of 6100 is 6.

Exercise 1.3
Question 1.

Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating, repeating decimal.

Since, this decimal has finite number of digits

∴ it is terminating

Question 2.

Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating, repeating decimal.

Since, this decimal has finite number of digits

∴ it is terminating.

Question 3.

Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating, repeating decimal.

Since, this decimal has finite number of digits

∴ it is terminating.

Question 4.

Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating, repeating decimal.

Since the decimal continues endlessly, it is non-terminating and repeating.

∴ it is non-terminating and repeating.

Question 5.

Write the following rational numbers in their decimal form and also state which are terminating and which have non-terminating, repeating decimal.

Since, this decimal has finite number of digits

∴ it is terminating.

Question 6.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 13 and 3125 are co-prime.

Now, We have to write the denominator 3125 in the form of 2n5m

where, n and m are the non-negative numbers.

3125 = 5 × 5 × 5 × 5 × 5 = 55

3125 ⇒ 1325 = 1 × 55 = 20 × 55

∴ denominator is of the form 2n5m where, n = 0 and m = 5

Thus, is a Terminating decimal

Question 7.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 11 and 12 are co-prime.

Now, We have to write the denominator 12 in the form of 2n5m where, n and m are the non-negative numbers.

12 = 2×2×3

12 = 22×3

∴ denominator is not of the form 2n5m where, n = 2 and m = 0

Thus, is Non-terminating and repeating decimal

Question 8.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 64 and 455 are co-prime.

Now, We have to write the denominator 455 in the form of 2n5m where, n and m are the non-negative numbers.

455 = 5×7×13

∴ denominator is not of the form 2n5m where, n = 0 and m = 1

Thus, is Non-terminating and repeating decimal

Question 9.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 3 and 320 are co-prime.

Now, We have to write the denominator 320 in the form of 2n5m where, n and m are the non-negative numbers.

320 = 2×2×2×2×2×2×5 = 2⁶×5

∴ denominator is in the form 2n5m where, n = 6 and m = 1

Thus, is terminating decimal.

Question 10.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 29 and 343 are co-prime.

Now, We have to write the denominator 343 in the form of 2n5m where, n and m are the non-negative numbers.

343 = 7×7×7 = 73

∴ denominator is not of the form 2n5m where, n = 0 and m = 0

Thus, is Non-terminating and repeating decimal

Thus, is Non-terminating decimal.

Question 11.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 23 and 23 52 are co-prime.

Now, we have to write the denominator 22 57 75 in the form of 2n5m where, n and m are the non-negative numbers.

23 52

∴ denominator is in the form 2n5m where, n = 3 and m = 2

Thus, is terminating decimal

Question 12.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 23 and 22 57 75 are co-prime.

Now, we have to write the denominator 22 57 75 in the form of 2n5m where, n and m are the non-negative numbers.

23 52

22×5⁷×7⁵

∴ denominator is not of the form 2n5m where, n = 2 and m = 7. Due to one more factor it is not in the form

Thus, is Non-terminating decimal.

Question 13.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 2 and 5 are co-prime.

Now, we have to write the denominator 5 in the form of 2n5m where, n and m are the non-negative numbers.

20 5

∴ denominator is in the form 2n5m where, n = 0 and m = 1

Thus, is Terminating decimal.

Question 14.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 7 and 10 are co-prime.

Now, we have to write the denominator 10 in the form of 2n5m where, n and m are the non-negative numbers.

10 = 2×5

∴ denominator is in the form 2n5m where, n = 1 and m = 1

Thus, is terminating decimal.

Question 15.

Without performing division, state whether the following rational numbers will have a terminating decimal form or non-terminating, repeating decimal form.

⇒ 11 and 30 are co-prime.

Now, We have to write the denominator 30 in the form of 2n5m where, n and m are the non-negative numbers.

30 = 2×3×5

∴ denominator is not of the form 2n5m where, n = 1 and m = 1. Due to one more factor it is not in the form

Thus, is Non-terminating and repeating decimal.

Question 16.

Write the following rationales in decimal form using Theorem 1.1.

(i)

(ii)

(iii)

(iv)

(v)

According to Euclid’s Division Algorithm,

For any two positive integers, ‘a’ and ‘b’, there exists a unique pair of integers ‘q’ and ‘r’ which satisfy the relation:

a = bq + r , 0 ≤ r ≤ b

(i)

(ii)

(iii)

(iv)

(v)

Question 17.

The decimal form of some real numbers is given below. In each case, decide whether the number is rational or not. If it is rational, and expressed in form what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000….

(iii) 43.

(i) 43.123456789

43.123456789 is terminating.

So, it would be a rational number

Hence, 43.123456789 is now in the form of .

And the prime factors of q are in terms of 2 and 5.

(ii) 0.120120012000120000….

0.120120012000120000…. is non-terminating and non-repeating.

So, it is not a rational number.

(iii) 43.

43. is non- terminating but terminating.

So, it would be a rational number.

In a non-terminating, repeating expansion of

q will have factors other than 2 or 5.

Exercise 1.4
Question 1.

Prove that the following are irrational.

Let be rational. Then, there exists positive co-primes a and b such that

is rational as a and b are integers

∴√2 is rational which contradicts to the fact that √2 is irrational.

Hence, our assumption is false and is irrational.

Question 2.

Prove that the following are irrational.

Let us suppose that is rational.

Let be rational equal to , where a and b are integers and a≠0 and b≠0 Then,

[squaring both sides]

Since a and b are integers, is rational. So, √3 is rational

Now, this contradicts the fact that √3 is rational.

Hence √3 + √5 is irrational.

Question 3.

Prove that the following are irrational.

6 + √2

Let 6 + √2 be a rational number equal to , where a,b are positive co-primes. Then,

6 + √2 =

⇒ √2

Since a and b are integers, is also rational and hence, √2should be rational. This is contracdicts the fact that√2is irrational. Therefore , our assumption is false and hence, 6 + √2 is irrational.

Question 4.

Prove that the following are irrational.

√5

Let take √5as rational number equal to , where a, b are positive co-primes. Then,

⇒ √5 b = a

⇒ 5b2 = a2[squaring both sides] …I

Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p divides a2 then it will divide a also.

∴ a = 5c

Put value of a in Eq. I, we get

5b2 = (5c)2

⇒ 5b2 = 25c2

[divide by 25 both sides]

Using same theorem we get that b will divide by 5 and we have already get that a is divided by 5. This contradicts our assumption.

Hence, √5 is irrational.

Question 5.

Prove that the following are irrational.

3 + 2√5

Let us assume on the contrary that 3 + 2√5 is rational. Then, there exist co-prime positive integers a and b such that

3 + 2√5

is rational [∵ a,b are integers ∴ is a rational]

This contradicts the fact that is irrational. So, our supposition is incorrect.

Hence, is an irrational number.

Question 6.

Prove that is irrational, where p, q are primes.

Let √p + √qbe rational

⇒ √p + √q [where a, b are co-primes and integers]

Squaring both sides

⇒ 2√pq = a^2/b^2 -p-q

….I

∴ from Eq. I our assumption contradicts here, because p is rational

Hence, √p + √q is irrational number.

Exercise 1.5
Question 1.

Determine the value of the following.

log255

The logarithmic form of log255

Using property of logarithmic, logb x = and loga ax = x

=

Question 2.

Determine the value of the following.

log813

The logarithmic form of log813

Using property of logarithmic, logb x = and loga ax = x

=

Question 3.

Determine the value of the following.

The logarithmic form of log2

Using property of logarithmic,

= log2 1 – log216

= 0 – log224

= 0 – 4 = -4

Question 4.

Determine the value of the following.

log71

The logarithmic form of log7 1

= 0

Question 5.

Determine the value of the following.

The logarithmic form of logx √x

Using the property of logarithmic, loga ax = x

= logx x1/2

=

Question 6.

Determine the value of the following.

The Logarithmic form of log2512

Using the property of logarithmic, loga ax = x

= log229

= 9

Question 7.

Determine the value of the following.

log100.01

The Logarithmic form of log100.01

= log10

Using property of logarithmic, and loga ax = x

= log101 – log10100

= 0 – log10102

= 0 – 2 = -2

Question 8.

Determine the value of the following.

The Logarithmic form of

Using the property of logarithmic, loga ax = x

=

=

= 3

Question 9.

Determine the value of the following.

22 + log23

The Logarithmic form of 22 + log23

Using property of logarithmic, am + n = am × an and alogax = x

= 22 + log23

= 22 × 2log23

= 4 × 3

= 12

Question 10.

Write each of the following expressions as log N. Determine the value of N. (You can assume the base is 10, but the results are identical which ever base is used).
(i) log2 + log5

(ii) log 16 – log 2

(iii) 3 log 4

(iv) 2 log 3 – 3 log 2

(v) log 243 + log 1

(vi) log 10 + 2 log 3 – log 2

Some basic logarithmic formulas are

1. alogab = b
2. loga 1 = 0
3. loga a = 1
4. loga(x · y) = logax + logay
5. loga xy = logax - logay
6. loga 1x = -logax
7. loga xp = p logax
8. logak x = 1k loga x, for k ≠ 0

I. log 2 + log 5

using the property of logarithm, loga xy = loga x + loga y

= log(2 × 5)

= log 10

N = 10

II. log 16 – log 2

using the property of logarithm,

and

= log (16/2)
= log 8
N = 8

III. 3 log 4

Using the property of logarithm,

log (4)3
= log 64
N = 64

IV. 2 log 3 - 3 log 2

using the property of logarithm,

and

= log 32-log 23

=

=

N = 9/8

V.
log 243 + log 1
Using log a + log b = log ab
= log 243(1)
= log 243

VI.
log 10 + 2 log 3 - log 2

using the property of logarithm, loga xy = loga x + loga y

= log 5 + log 2 + log 32 - log 2

= log 5 + log 9

= log (5×9)

= log 45

Question 11.

Evaluate each of the following in terms of x and y, if it is given x = log23 and y = log25

(i) log2 15 (ii) log27.5

(iii) log260 (iv) log26750

I. log215

⇒ log25 + log23 〗[∵]

⇒ x + y

II. log27.5

⇒ log2

⇒ log215 - log22 [∵]

⇒ log25 + log23-log22[∵]

⇒ x + y-1

III.

⇒ log212 + log25 [∵]

⇒ log24 + log23 + log25 [∵]

⇒ 2log22 + log23 + log25

⇒ 2(1) + x + y = x + y + 2

IV. log26750

⇒ log254 + log2125〗

⇒ log227 + log22 + log2125 〗 〗

⇒ log233 + log253 + log22

⇒ 3 log23 + 3 log25 + log22

⇒ 3x + 3y + 1

Question 12.

Expand the following.

(i) log1000

(ii)

(iii) logx2y3z4

(iv)

(v)

i.

ii. =

⇒ log27 - log54

⇒ 7log2 - 4 log5

iii. logx2 y3 z4

⇒ logx2 + + logy3 + logz4

⇒ 2 logx + 3 logy + 4 logz

iv. logp2 q3-logr

⇒ logp2 + log q3-log r

⇒ 2 log p + 3 log q-log r

v. =

{3log x + 2 log y} = log x + log y

Question 13.

If x2 + y2 = 25xy, then prove that 2 log(x + y) = 3log3 + logx + logy.

x2 + y2 = 25xy

x2 + y2 + 2xy = 27xy

(x + y)2 = 27xy

Taking log both sides

log (x + y)2 = log 27xy

2 log (x + y) = log 27 + log x + log y

2 log(x + y) = log 33 + log x + log y

2 log (x + y) = 3 log 3 + log x + log y

Hence proved.

Question 14.

If then find the value of

[∵] [∵]

Remove log from both sides

= xy

⇒ x2 + y2 + 2xy = 9xy

⇒ x2 + y2 = 7xy

Divide by both sides

Question 15.

If (2.3) x = (0.23) y = 1000 then find the value of

2.3x = 0.23y = 1000

Consider 2.3x = 1000

⇒ log 2.3x = log103

⇒ x log2.3 = 3 log 10

⇒ log2.3 = [∵ log10 = 1] …..I

Consider 0.23y = 1000

⇒ log 0.23y = log 1000

⇒ y log0.23 = log 103

⇒ y log 0.23 = 3 log 10

⇒ log 0.23 = ……..II

Subtract eq. II from I

Question 16.

If then find the value of x.

2(x + 1) = 3(1-x)

Taking log base both sides

log(2(x + 1)) = log(3(1-x))

(x + 1) log 2 = (1 - x) log 3

x log 2 + x log 3 = log 3 - log 2

x(log 2 + log 3) = log 3 - log 2

x = (log 3 - log 2)/(log 2 + log 3)

Question 17.

Is

Assume that log 2 is rational, that is,

…(1)

where p, q are integers.

Since, and therefore, p<q

From Eq. 1,

2q = (2×5)p

2(q-p) = 5p

Where p-q is an integer greater than 0.

Now, it can be seen that the L.H.S is even and the R.H.S is odd.

Hence, there is contradiction and log 2 is irrational.

Question 18.

Let us assume log 100 is rational

log 100 = log10 100

log10 102 = 2 log10 10 = 2

As 2 is rational number, ∴ log 100 is also rational.