ML Aggarwal Solutions for Class 9 Maths Chapter 9 – Logarithms are provided here to help students prepare for their exams and score well. This chapter mainly deals with problems based on logarithms. Solutions are prepared by our experienced faculty team at BYJUâ€™S, which are created for the purpose of clarifying studentâ€™s doubts, as per their convenience. These solutions also provide guidance to students for solving problems confidently, which in turn, helps in improving their problem-solving skills. This is very important from the examination point of view. For students wishing to learn the right steps of solving such problems, the ML Aggarwal Solutions is made available in the pdf format. Students can easily download the pdf from the links given below.

Chapter 9 – Logarithms contains two exercises and the ML Aggarwal Class 9 Solutions present in this page provide solutions to questions related to each exercise present in this chapter.

## Download the Pdf of ML Aggarwal Solutions for Class 9 Maths Chapter 9 – Logarithms

### Access answers to ML Aggarwal Solutions for Class 9 Maths Chapter 9 – Logarithms

EXERCISE 9.1

**1. Convert the following to logarithmic form: (i) 5 ^{2}Â = 25(ii) a^{5}Â = 64(iii) 7^{x}Â = 100(iv) 9^{o} = 1(v) 6^{1}Â = 6(vi) 3^{-2}Â =Â 1/9(vii) 10^{-2}Â = 0.01(viii) (81)^{3/4}Â = 27**

**Solution:**

**(i)** 5^{2}Â = 25

Let us apply log, we get

Log_{5} 25 = 2

**(ii)** a^{5}Â = 64

Let us apply log, we get

Log_{a} 64 = 5

**(iii)** 7^{x}Â = 100

Let us apply log, we get

Log_{7} 100 = x

**(iv)** 9^{o} = 1

Let us apply log, we get

Log_{9} 1 = 0

**(v)** 6^{1}Â = 6

Let us apply log, we get

Log_{6} 6 = 1

**(vi)** 3^{-2}Â =Â 1/9

Let us apply log, we get

Log_{3} 1/9 = -2

**(vii)** 10^{-2}Â = 0.01

Let us apply log, we get

Log_{10} 0.01 = -2

**(viii)** (81)^{3/4}Â = 27

Let us apply log, we get

Log_{81} 27 = Â¾

**2. Convert the following into exponential form: **

**(i) log _{2}Â 32 = 5**

**(ii) log**

_{3}Â 81=4**(iii) log**

_{3 }1/3 = -1**(iv) log**

_{3}Â 4 =Â 2/3**(v) log**

_{8}Â 32 =Â 5/3**(vi) log**

_{10}Â (0.001) = -3**(vii) log**

_{2}Â 0.25 = -2**(viii) log**

_{a}Â (1/a) = -1**Solution:**

**(i) **log_{2}Â 32 = 5

The exponential form of the expression is

2^{5} = 32

**(ii)** log_{3}Â 81=4

The exponential form of the expression is

3^{4} = 81

**(iii)** log_{3 }1/3 = -1

The exponential form of the expression is

3^{-1} = 1/3

**(iv)** log_{3}Â 4 =Â 2/3

The exponential form of the expression is

(8)^{2/3} = 4

**(v)** log_{8}Â 32 =Â 5/3

The exponential form of the expression is

(8)^{5/3} = 32

**(vi)** log_{10}Â (0.001) = -3

The exponential form of the expression is

10^{-3} = 0.001

**(vii)** log_{2}Â 0.25 = -2

The exponential form of the expression is

2^{-2} = 0.25

**(viii)** log_{a}Â (1/a) = -1

The exponential form of the expression is

a^{-1} = 1/a

**3.** **By converting to exponential form, find the values of: **

**(i) log _{2}Â 16**

**(ii) log**

_{5}Â 125**(iii) log**

_{4}Â 8**(iv) log**

_{9}Â 27**(v) log**

_{10}(.01)**(vi) log**

_{7 }1/7**(vii) log.**

_{5}Â 256**(viii) log**

_{2}Â 0.25**Solution:**

**(i) **log_{2}Â 16

Let us consider log_{2}Â 16 = x

So,

(2)^{x} = 16

= 2 Ã— 2 Ã— 2 Ã— 2

2^{x} = 2^{4}

By comparing the powers,

x = 4

**(ii)** log_{5}Â 125

Let us consider log_{5}Â 125 = x

So,

(5)^{x} = 125

= 5 Ã— 5 Ã— 5

5^{x} = 5^{3}

By comparing the powers,

x = 3

**(iii)** log_{4}Â 8

Let us consider log_{4}Â 8 = x

So,

(4)^{x} = 8

(2 Ã— 2)^{x} = 2 Ã— 2 Ã— 2

2^{2x} = 2^{3}

By comparing the powers,

2x = 3

x = 3/2

**(iv)** log_{9}Â 27

Let us consider log_{9}Â 27= x

So,

(9)^{x} = 27

(3 Ã— 3)^{x} = 3 Ã— 3 Ã— 3

3^{2x} = 3^{3}

By comparing the powers,

2x = 3

x = 3/2

**(v)** log_{10 }(.01)

Let us consider log_{10 }(.01) = x

So,

(10)^{x} = 1/100

= 1/10 Ã— 1/10

10^{x} = 1/(10)^{2}

10^{x} = 10^{-2}

By comparing the powers,

x = -2

**(vi)** log_{7 }1/7

Let us consider log_{7 }1/7 = x

So,

(7)^{x} = 1/7

7^{x} = 7^{-1}

By comparing the powers,

x = -1

**(vii)** log._{5}Â 256

Let us consider log._{5}Â 256 = x

So,

(.5)^{x} = 256

(5/10)^{x} = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

(1/2)^{x} = 2^{8}

(2)^{-x} = 2^{8}

By comparing the powers,

-x = 8

x = -8

**(viii)** log_{2}Â 0.25

Let us consider log_{2}Â 0.25 = x

So,

(2)^{x} = 0.25

= 25/100

2^{x} = 1/4

2^{x} = (2)^{-2}

By comparing the powers,

x = -2

**4. Solve the following equations for x:**

**(i) log _{3}Â x = 2**

**(ii) log _{x}Â 25 = 2**

**(iii) log _{10}Â x = -2**

**(iv) log _{4}Â x = Â½ **

**(v) log _{x}Â 11 = 2.5**

**(vi) log _{x}Â Â¼ = -1**

**(vii) log _{81}Â x = 3/2**

**(viii) log _{9}Â x = 2.5**

**(ix) log _{4}Â x = -1.5**

**Solution:**

**(i)** log_{3}Â x = 2

Let us simplify the expression,

(3)^{2} = x

x = 9

**(ii) **log_{x}Â 25 = 2

Let us simplify the expression,

(x)^{2} = 25

= 5 Ã— 5

x^{2} = 5^{2}

Since the powers are same,

So,

x = 5

**(iii)** log_{10}Â x = -2

Let us simplify the expression,

(10)^{-2} = x

x = 1/(10)^{2}

= 1/100

x = 0.01

**(iv)** log_{4}Â x = Â½

Let us simplify the expression,

(4)^{1/2} = x

x = (2 Ã— 2)^{1/2}

= (2)^{2}Ã—^{1/2}

x = 2

**(v)** log_{x}Â 11 = 2.5

Let us simplify the expression,

(x)^{1} = 11

x = 11

**(vi)** log_{x}Â Â¼ = -1

Let us simplify the expression,

(x)^{-1} = Â¼

x^{-1} = 4^{-1}

Since the powers are same,

So,

x = 4

**(vii)** log_{81}Â x = 3/2

Let us simplify the expression,

(81)^{3/2} = x

x = 81^{3/2}

= (3^{4})^{3/2}

= 3^{4}Ã—^{3/2}

= 3^{6}

= 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

= 729

x = 729

**(viii)** log_{9}Â x = 2.5

log_{9}Â x = 5/2

Let us simplify the expression,

(9)^{5/2} = x

x = (3^{2})^{5/2}

= 3^{2}Ã—^{5/2}

= 3^{5}

= 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

= 234

x = 234

**(ix)** log_{4}Â x = -1.5

log_{4}Â x = -3/2

Let us simplify the expression,

(4)^{-3/2} = x

x = (2^{2})^{-3/2}

= 2^{2}Ã—^{-3/2}

= 2^{-3}

= 1/2^{3}

= 1/(2 Ã— 2 Ã— 2)

= 1/8

x = 1/8

**5. Given log _{10 }a = b, express 10^{2b-3}Â in terms of a.**

**Solution:**

Given:

log_{10 }a = b

(10)^{b} = a

Now,

10^{2b-3} = (10)^{2b} / (10)^{3}

= (10^{b})^{2} / (10 Ã— 10 Ã— 10)

Substitute the value of (10)^{b} = a, we get

= a^{2}/1000

**6.** **Given log _{10}Â x= a, log_{10}Â y = b and log_{10}Â z =c,**

**(i) write down 10**

^{2a-3}Â in terms of x.**(ii) write down 10**

^{3b-1}Â in terms of y.**(iii) if log**

_{10}Â P = 2a +Â b/2 â€“ 3c, express P in terms of x, y and z.**Solution:**

Given:

log_{10}Â x= a

- (10)
^{a}= x

log_{10}Â y = b

- (10)
^{b}= y

log_{10}Â z =c

- (10)
^{c}= z

**(i) **Write down 10^{2a-3}Â in terms of x.

10^{2a-3}Â = (10)^{2a} / (10)^{3}

= (10^{a})^{2} / (10 Ã— 10 Ã— 10)

Substitute the value of (10)^{a} = x, we get

= x^{2}/1000

**(ii) **Write down 10^{3b-1}Â in terms of y.

10^{3b-1}Â = (10)^{3b} / (10)^{1}

= (10^{b})^{3} / (10)

Substitute the value of (10)^{b} = y, we get

= y^{3}/10

**(iii) **If log_{10}Â P = 2a +Â b/2 â€“ 3c, express P in terms of x, y and z.

we know that,

(10)^{a} = x

(10)^{b} = y

(10)^{c} = z

By substituting the values

log_{10}Â P = 2a +Â b/2 â€“ 3c

= 2 log_{10} x + Â½ log_{10} y â€“ 3 log_{10} z

= log_{10} x^{2} + log_{10} y^{1/2 }– log_{10} z^{3}

= log_{10} (x^{2} + y^{1/2}) – log_{10} z^{3}

= log_{10} [(x^{2}**âˆš**y)/z^{3}]

P = (x^{2}**âˆš**y)/z^{3}

**7. If log _{10}x = a and log_{10}y = b, find the value of xy.**

**Solution:**

Given:

log_{10}x = a

(10)^{a} = x

log_{10}y = b

(10)^{b} = y

Then,

xy = (10)^{a} Ã— (10)^{b}

= (10)^{a+b}

**8. Given log _{10}Â a = m and log_{10}Â b = n, express a^{3}/b^{2}Â in terms of m and n.**

**Solution:**

Given:

log_{10}Â a = m

(10)^{m} = a

log_{10}Â b = n

(10)^{n} = b

So,

a^{3}/b^{2}Â = (10^{m})^{3} / (10^{n})^{2}

= 10^{3m} / 10^{2n}

= 10^{3m â€“ 2n}

**9.** **Given log _{10}a= 2a and log_{10}y = â€“b/2**

**(i) write 10**

^{a}Â in terms of x.**(ii) write 10**

^{2b+1}Â in terms of y.**(iii) if log**

_{10}P= 3a – 2b, express P in terms of x and y.**Solution:**

Given:

log_{10}a= 2a

(10)^{2a} = a

log_{10}y = â€“b/2

(10)^{-b/2} = y

**(i) **Write 10^{a}Â in terms of x.

10^{a} = (10^{2a})^{1/2}

= (x)^{1/2}

= **âˆš**x

**(ii) **Write 10^{2b+1}Â in terms of y.

10^{2b+1} = 10^{2b} Ã— 10^{1}

= 10^{4(b/2)} Ã— 10^{1}

= (10^{b/2})^{4} Ã— 10^{1}

= y^{4} Ã— 10^{1}

= 10y^{4}

**(iii) **If log_{10}P= 3a – 2b, express P in terms of x and y.

log_{10}P= 3a – 2b

Substitute the values,

log_{10}P= 3/2 (2a) â€“ 4(b/2)

= 3/2 (log_{10} x) â€“ 4 (log_{10} y)

= (log_{10} x)^{3/2} – (log_{10} y)^{4}

= log_{10} [(x^{3/2}) / y^{4}]

P = (x^{3/2}) / y^{4}

**10.** **If log _{2}Â y = x and log_{3}Â z = x, find 72^{x}Â in terms of y and z.**

**Solution:**

Given:

log_{2}Â y = x

2^{x} = y

log_{3}Â z = x

3^{x} = z

So,

72^{x} = (2 Ã— 2 Ã— 2 Ã— 3 Ã— 3)^{x}

= (2^{3} Ã— 3^{2})^{x}

= 2^{3x} Ã— 3^{2x}

= (2^{x})^{3} Ã— (3^{x})^{2}

= y^{3} Ã— z^{2}

= y^{3}z^{2}

**11.** **If log _{2}Â x = a and log_{5}y = a, write 100^{2a-1}Â in terms of x and y.**

**Solution:**

Given:

log_{2}Â x = a

2^{a} = x

log_{5}y = a

5^{a} = y

So,

100^{2a-1} = (2 Ã— 2 Ã— 5 Ã— 5)^{2a-1}

= (2^{2} Ã— 5^{2})^{2a-1}

= 2^{4a-2} Ã— 5^{a-2}

= (2^{4a})/2^{2} Ã— (5^{4a})/5^{2}

= [(2^{a})^{4} Ã— (5^{a})^{4}] / (4Ã—25)

= (x^{4}y^{4})/100

EXERCISE 9.2

**1. Simplify the following:**

**(i) log a ^{3} â€“ log a^{2} **

**(ii) log a ^{3} Ã· log a^{2}**

**(iii) log 4/log 2**

**(iv) (log 8 log 9)/log 27**

**(v) log 27/log âˆš3**

**(vi) (log 9 â€“ log 3)/log 27**

**Solution:**

**(i) **log a^{3} â€“ log a^{2}** **

By using Quotient law,

log a^{3} â€“ log a^{2} = log (a^{3}/a^{2})

= log a

**(ii) **log a^{3} Ã· log a^{2}

By using power law,

log a^{3} Ã· log a^{2} = 3log a Ã· 2 log a

= 3log a / 2log a

= 3/2

**(iii) **log 4/log 2

Let us simplify the expression,

log 4/log 2 = log(2Ã—2)/log 2

By using power law,

= 2 log 2/log 2

= 2

**(iv) **(log 8 log 9)/log 27

Let us simplify the expression,

(log 8 log 9)/log 27 = (log 2^{3}. log 3^{2})/log 3^{3}

By using power law,

= [(3 log 2).(2 log 3)]/(3 log 3)

= [(log 2).2] / 1

= 2 log 2

= log 2^{2}

= log 4

**(v) **log 27/log âˆš3

Let us simplify the expression,

log 27/log âˆš3 = log(3Ã—3Ã—3)/log(3)^{1/2}

= log 3^{3}/log 3^{1/2}

By using power law

= 3log 3/((1/2)log 3)

= (3Ã—2)/1 (log 3/log 3)

= (6) (1)

= 6

**(vi) **(log 9 â€“ log 3)/log 27

Let us simplify the expression,

(log 9 â€“ log 3)/log 27 = [log(3Ã—3) â€“ log 3] / log(3Ã—3Ã—3)

= [log 3^{2} â€“ log 3] / log 3^{3 }

By using power law

= [2 log 3 â€“ log 3]/3log 3

= log 3/3log 3

= 1/3

**2. Evaluate the following:**

**(i) log (10 Ã· âˆ›10)**

**(ii) 2 + Â½ log(10 ^{-3})**

**(iii) 2 log 5 + log 8 â€“ Â½ log 4**

**(iv) 2 log 10 ^{3} + 3 log 10^{-2} – 1/3 log 5^{-3} + Â½ log 4**

**(v) 2 log 2 + log 5 â€“ Â½ log 36 â€“ log 1/30**

**(vi) 2 log 5 + log 3 + 3 log 2 â€“ Â½ log 36 â€“ 2 log 10**

**(vii) log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80**

**(viii) 2 log _{10} 5 + log_{10} 8 â€“ Â½ log_{10} 4**

**Solution:**

**(i) **log (10 Ã· âˆ›10)

Let us simplify the expression,

log (10 Ã· âˆ›10) = log (10 Ã· 10^{1/3})

= log (10^{1 â€“ 1/3})

= log (10^{2/3})

= 2/3 log 10

= 2/3 (1)

= 2/3

**(ii) **2 + Â½ log(10^{-3})

Let us simplify the expression,

2 + Â½ log(10^{-3}) = 2 + Â½ Ã— (-3) log 10

= 2 â€“ 3/2 log 10

= 2 â€“ 3/2 (1)

= 2 â€“ 3/2

= (4-3)/2

= Â½

**(iii) **2 log 5 + log 8 â€“ Â½ log 4

Let us simplify the expression,

2 log 5 + log 8 â€“ Â½ log 4 = log 5^{2} + log 8 â€“ Â½ log 2^{2}

= log 25 + log 8 â€“ Â½ 2log 2

= log 25 + log 8 â€“ log 2

= log (25Ã—8)/2

= log (25Ã—4)

= log 100

= log 10^{2}

= 2 log 10

= 2 (1)

= 2

**(iv) **2 log 10^{3} + 3 log 10^{-2} – 1/3 log 5^{-3} + Â½ log 4

Let us simplify the expression,

2 log 10^{3} + 3 log 10^{-2} – 1/3 log 5^{-3} + Â½ log 4 = 2Ã—3 log 10 + 3(-2)log 10 â€“ 1/3 (-3) log 5 + Â½ log 2^{2}

= 6 log 10 â€“ 6 log 10 + log 5 + Â½ 2 log 2

= 6 log 10 â€“ 6 log 10 + log 5 + log 2

= 0 + log 5 + log 2

= log (5Ã—2)

= log 10

= 1

**(v) **2 log 2 + log 5 â€“ Â½ log 36 â€“ log 1/30

Let us simplify the expression,

2 log 2 + log 5 â€“ Â½ log 36 â€“ log 1/30 = log 2^{2} + log 5 â€“ Â½ log 6^{2} â€“ log (1/30)

= log 4 + log 5 â€“ log 6 â€“ log 1/30

= log 4 + log 5 â€“ log 6 â€“ (log 1 â€“ log 30)

= log 4 + log 5 â€“ log 6 â€“ log 1 + log 30

= log 4 + log 5 + log 30 â€“ (log 6 + log 1)

= log (4Ã—5Ã—30) â€“ log (6Ã—1)

= log (4Ã—5Ã—30)/(6Ã—1)

= log (4Ã—5Ã—5)

= log 100

= log 10^{2}

= 2 log 10

= 2 (1)

= 2

**(vi) **2 log 5 + log 3 + 3 log 2 â€“ Â½ log 36 â€“ 2 log 10

Let us simplify the expression,

2log 5 + log 3 + 3log 2 â€“ Â½ log 36 â€“ 2log 10** = **log 5^{2} + log 3 + log 2^{3} â€“ Â½ log 6^{2} â€“ log 10^{2}

= log 25 + log 3 + log 8 â€“ log 6 â€“ log 100

= log (25Ã—3Ã—8) â€“ log(6Ã—100)

= log (25Ã—3Ã—8)/ (6Ã—100)

= log (1Ã—3Ã—8) / (6Ã—4)

= log 24/24

= log 1

= 0

**(vii) **log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80

Let us simplify the expression,

log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80 = log 2 + 16(log 16 â€“ log 15) + 12(log 25 â€“ log 24) + 7(log 81 â€“ log 80)

= log 2 + 16 (log 2^{4} â€“ log (3Ã—5)) + 12 (log 5^{2} â€“ log (3Ã—2Ã—2Ã—2)) + 7 (log (3Ã—3Ã—3Ã—3) â€“ log (2^{4} Ã— 5))

= log 2 + 16(4log 2 â€“ (log 3 + log 5)) + 12 (2log 5 â€“ log (3Ã—2^{3})) + 7 (log 3^{4} â€“ (log 2^{4} + log 5))

= log 2 + 16 (4log 2 â€“ log 3 â€“ log 5) + 12 (2log 5 â€“ (log 3 + 3log 2)) + 7 (4log 3 â€“ 4log 2 â€“ log 5)

= log 2 + 64log 2 â€“ 16log 3 â€“ 16log 5 + 24log 5 â€“ 12log 3 â€“ 36log 2 + 28log 3 â€“ 28log 2 â€“ 7log 5

= (log 2 + 64log 2 â€“ 36log 2 â€“ 28log 2) + (-16log 3 â€“ 12log 3 + 28log 3) + (-16log 5 + 24log 5 â€“ 7log 5)

= (65log 2 â€“ 64log 2) + (-28log 3 + 28log 3) + (-23log 5 + 24log 5)

= log 2 + 0 + log 5

= log (2Ã—5)

= log 10

= 1

**(viii) **2 log_{10} 5 + log_{10} 8 â€“ Â½ log_{10} 4

Let us simplify the expression,

2 log_{10} 5 + log_{10} 8 â€“ Â½ log_{10} 4 = log_{10} 5^{2} + log_{10} 8 â€“ log_{10} 4^{1/2}

= log_{10} 25 + log_{10} 8 â€“ log_{10} (2)^{2}Ã—^{1/2}

= log_{10} 25 + log_{10} 8 â€“ log_{10} 2

= log_{10} [(25Ã—8)/2]

= log_{10} (25Ã—4)

= log_{10} 100

= log_{10} 10^{2}

= 2 log_{10} 10

= 2 (1)

= 2

**3. Express each of the following as a single logarithm:**

**(i) 2 log 3 â€“ Â½ log 16 + log 12**

**(ii) 2 log _{10} 5 â€“ log_{10} 2 + 3 log_{10} 4 + 1**

**(iii) Â½ log 36 + 2 log 8 â€“ log 1.5**

**(iv) Â½ log 25 â€“ 2 log 3 + 1**

**(v) Â½ log 9 + 2 log 3 â€“ log 6 + log 2 – 2**

**Solution:**

**(i) **2 log 3 â€“ Â½ log 16 + log 12

Let us simplify the expression into single logarithm,

2 log 3 â€“ Â½ log 16 + log 12 = 2 log 3 â€“ Â½ log 4^{2} + log 12

= 2 log 3 â€“ log 4 + log 12

= log 3^{2} â€“ log 4 + log 12

= log 9 – log 4 + log 12

= log (9Ã—12)/4

= log (9Ã—3)

= log 27

**(ii) **2 log_{10} 5 â€“ log_{10} 2 + 3 log_{10} 4 + 1

Let us simplify the expression into single logarithm,

2 log_{10} 5 â€“ log_{10} 2 + 3 log_{10} 4 + 1 = log_{10} 5^{2} â€“ log_{10} 2 + log_{10} 4^{3} + log_{10} 10

= log_{10} 25 â€“ log_{10} 2 + log_{10} 64 + log_{10} 10

= log_{10} (25Ã—64Ã—10) â€“ log_{10} 2

= log_{10} (16000) â€“ log_{10} 2

= log_{10} (16000/2)

= log_{10} 8000

**(iii) **Â½ log 36 + 2 log 8 â€“ log 1.5

Let us simplify the expression into single logarithm,

Â½ log 36 + 2 log 8 â€“ log 1.5 = log 36^{1/2} + log 8^{2} â€“ log 1.5

= log (6)^{2}Ã—^{1/2} + log 64 â€“ log 1.5

= log 6 + log 64 â€“ log (15/10)

= log 6 + log 64 â€“ (log 15 â€“ log 10)

= log (6Ã—64) â€“ log 15 + log 10

= log (6Ã—64Ã—10) â€“ log 15

= log [(6Ã—64Ã—10)/15]

= log (4Ã—64)

= log 256

**(iv) **Â½ log 25 â€“ 2 log 3 + 1

Let us simplify the expression into single logarithm,

Â½ log 25 â€“ 2 log 3 + 1 = log 25^{1/2} â€“ log 3^{2} + log 10

= log (5)^{2}Ã—^{1/2} â€“ log 9 + log 10

= log 5 â€“ log 9 + log 10

= log (5Ã—10) â€“ log 9

= log ((5Ã—10)/9)

= log 50/9

**(v) **Â½ log 9 + 2 log 3 â€“ log 6 + log 2 – 2

Let us simplify the expression into single logarithm,

Â½ log 9 + 2 log 3 â€“ log 6 + log 2 â€“ 2 = log 9^{1/2} + log 3^{2} â€“ log 6 + log 2 â€“ log 100

= log 3^{2}Ã—^{1/2} + log 9 â€“ log 6 + log 2 â€“ log 100

= log 3 + log 9 â€“ log 6 + log 2 â€“ log 100

= log [(3Ã—9Ã—2)/(6Ã—100)]

= log 9/100

**4. Prove the following:**

**(i) log _{10} 4 Ã· log_{10} 2 = log_{3} 9**

**(ii) log _{10} 25 + log_{10} 4 = log_{5} 25**

**Solution:**

**(i) **log_{10} 4 Ã· log_{10} 2 = log_{3} 9

Let us consider LHS, log_{10} 4 Ã· log_{10} 2

log_{10} 4 Ã· log_{10} 2 = log_{10} 2^{2} Ã· log_{10} 2

= 2 log_{10} 2 Ã· log_{10} 2

= 2 log_{10} 2/ log_{10} 2

= 2 (1)

= 2

Now let us consider RHS,

log_{3} 9 = log_{3} 3^{2}

= 2 log_{3} 3

= 2(1)

= 2

âˆ´ LHS = RHS

Hence proved.

**(ii) **log_{10} 25 + log_{10} 4 = log_{5} 25

Let us consider LHS, log_{10} 25 + log_{10} 4

log_{10} 25 + log_{10} 4 = log_{10} (25Ã—4)

= log_{10} 100

= log_{10} 10^{2}

= 2 log_{10} 10

= 2(1)

= 2

Now, let us consider RHS,

log_{5} 25 = log_{5} 5^{2}

= 2 log_{5} 5

= 2 (1)

= 2

âˆ´ LHS = RHS

Hence proved.

**5. If x = (100) ^{a}, y = (10000)^{b}Â and z = (10)^{c}, express log [(10âˆšy)/x^{2}z^{3}] in terms of a, b, c.**

**Solution:**

Given:

x = (100)^{a} = (10^{2})^{a} = 10^{2a}

y = (10000)^{b}Â = (10^{4})^{b} = 10^{4b}

z = (10)^{c}

It is given that, log [(10âˆšy)/x^{2}z^{3}]

log [(10âˆšy)/x^{2}z^{3}] = (log 10 + log âˆšy) â€“ (log x^{2} + log z^{3})

= (1 + log(y)^{1/2}) â€“ (log x^{2} + log z^{3}) [we know that, log 10 = 1]

= (1 + Â½ log y) â€“ (2 log x + 3 log z)

Now substitute the values of x, y, z, we get

= (1 + Â½ log 10^{4b}) â€“ (2 log 10^{2a} + 3 log 10^{c})

= (1 + Â½ 4b log 10) â€“ (2Ã—2a log 10 + 3Ã—c log 10)

= (1 + Â½ 4b) â€“ (2Ã—2a + 3c) [Since, log 10 = 1]

= (1 + 2b) â€“ (4a + 3c)

= 1 + 2b â€“ 4a â€“ 3c

**6.** **If a = log _{10}x, find the following in terms of a :**

**(i) x**

**(ii) log**

_{10}^{5}âˆšx^{2}**(iii) log**

_{10}5x**Solution:**

Given:

a = log_{10}x

**(i)** x

10^{a} = x

âˆ´ x = 10^{a}

**(ii) **log_{10} ^{5}âˆšx^{2}

log_{10} ^{5}âˆšx^{2} = log_{10} (x^{2})^{1/5}

= log_{10} (x)^{2/5}

= 2/5 log_{10} x

= 2/5 (a)

= 2a/5

**(iii) **log_{10} 5x

x = (10)^{a}

= log_{10} 5x

= log_{10} 5(10)^{a}

= log_{10} 5 + log_{10} 10

= log_{10} 5 + a(1)

= a + log_{10} 5

**7. If a =log 2/3, b = log 3/5 and c = 2 log âˆš(5/2). Find the value of **

**(i) a + b + c **

**(ii) 5 ^{a+b+c} **

**Solution:**

Given:

a =log 2/3

b = log 3/5

c = 2 log âˆš(5/2)

(i) a + b + c

Let us substitute the given values, we get

a + b + c = log 2/3 + log 3/5 + 2 log âˆš(5/2)

= (log 2 â€“ log 3) + (log 3 â€“ log 5) + 2 log (5/2)^{1/2}

= log 2 â€“ log 3 + log 3 â€“ log 5 + 2 Ã— Â½ (log 5 â€“ log 2)

= log 2 â€“ log 3 + log 3 â€“ log 5 + log 5 â€“ log 2

= 0

**(ii) **5^{a+b+c}

5^{a+b+c} = 5^{0}

= 1

**8. If x = log 3/5, y = log 5/4 and z = 2 log âˆš3/2, find the value of **

**(i) x + y â€“ z **

**(ii) 3 ^{x+y-z}**

**Solution:**

Given:

x = log 3/5 = log 3 â€“ log 5

y = log 5/4 = log 5 â€“ log 4

z = 2 log âˆš3/2 = log (âˆš3/2)^{2} = log Â¾ = log 3 â€“ log 4

**(i) **x + y â€“ z** **

Let us substitute the given values, we get

x + y â€“ z = log 3 â€“ log 5 + log 5 â€“ log 4 â€“ (log 3 â€“ log 4)

= log 3 â€“ log 5 + log 5 â€“ log 4 â€“ log 3 + log 4

= 0

**(ii) **3^{x+y-z}

3^{x+y-z} = 3^{0}

= 1

**9.** **If x = log _{10}Â 12, y = log_{4}Â 2 **Ã—

**log**

_{10}Â 9 and z = log_{10}Â 0.4, find the values of**(i) x â€“ y â€“ z**

**(ii) 7**

^{x-y-z}**Solution:**

Given:

x = log_{10}Â 12

y = log_{4}Â 2 Ã— log_{10}Â 9

z = log_{10}Â 0.4

**(i) **x â€“ y â€“ z

Let us substitute the given values, we get

x â€“ y â€“ z = log_{10}Â 12 – log_{4}Â 2 Ã— log_{10}Â 9 – log_{10}Â 0.4

= log_{10} (3Ã—4) â€“ log_{4} 4^{1/2} Ã— log_{10}Â 3^{2} â€“ log_{10} 4/10

= log_{10} 3 + log_{10} 4 â€“ Â½ log_{4} 4 Ã— 2 log_{10} 3 â€“ (log_{10}^{ }4 â€“ log_{10} 10)

= log_{10} 3 + log_{10} 4 â€“ Â½ Ã— 1 Ã— 2 log_{10} 3 â€“ log_{10}^{ }4 + 1

= log_{10} 3 + log_{10} 4 – log_{10} 3 â€“ log_{10}^{ }4 + 1

= 1

**(ii) **7^{x-y-z}

7^{x-y-z} = 7^{1}

= 7

**10. If log V + log 3 = log Ï€ + log 4 + 3 log r, find V in terms of other quantities.Solution: **

Given:

log V + log 3 = log Ï€ + log 4 + 3 log r

Let us simplify the given expression to find V,

log (V Ã— 3) = log (Ï€ Ã— 4 Ã— r^{3})

log 3V = log 4Ï€r^{3}

3V = 4Ï€r^{3}

V = 4Ï€r^{3}/3

**11.** **Given 3 (log 5 â€“ log 3) â€“ (log 5 – 2 log 6) = 2 â€“ log n, find n.**

**Solution:**

Given:

3 (log 5 â€“ log 3) â€“ (log 5 – 2 log 6) = 2 â€“ log n

Let us simplify the given expression to find n,

3 log 5 â€“ 3 log 3 â€“ log 5 + 2 log 6 = 2 â€“ log n

2 log 5 â€“ 3 log 3 + 2 log 6 = 2 (1) â€“ log n

log 5^{2} â€“ log 3^{3} + log 6^{2} = 2 log 10 â€“ log n [Since, 1 = log 10]

log 25 â€“ log 27 + log 36 â€“ log 10^{2} = – log n

log n = – log 25 + log 27 â€“ log 36 + log 100

= (log 100 + log 27) â€“ (log 25 + log 36)

= log (100Ã—27) â€“ log (25Ã—36)

= log (100Ã—27)/(25Ã—36)

log n = log 3

n = 3

**12. Given that log _{10} y + 2 log_{10 }x = 2, express y in terms of x.**

Solution:

Given:

log_{10} y + 2 log_{10 }x = 2

Let us simplify the given expression,

log_{10} y + log_{10 }x^{2} = 2(1)

log_{10} y + log_{10 }x^{2} = 2 log_{10} 10

log_{10} (yÃ—x^{2}) = log_{10} 10^{2}

yx^{2} = 100

y = 100/x^{2}

**13.** **Express log _{10 }2 + 1 in the form log_{10}x.**

**Solution:**

Given:

log_{10 }2 + 1

Let us simplify the given expression,

log_{10 }2 + 1 = log_{10} 2 + log_{10} 10 [Since, 1 = log_{10} 10 ]

= log_{10} (2Ã—10)

= log_{10} 20

**14. If a ^{2} = log_{10} x, b^{2} = log_{10} y and a^{2}/2 â€“ b^{2}/3 = log_{10} z. Express z in terms of x and y.**

**Solution:**

Given:

a^{2} = log_{10} x

b^{2} = log_{10} y

a^{2}/2 â€“ b^{2}/3 = log_{10} z

Let us substitute the given values in the expression, we get

log_{10} x/2 – log_{10} y/3 = log_{10} z

log_{10} x^{1/2} â€“ log_{10} y^{1/3} = log_{10} z

log_{10} **âˆš**x â€“ log_{10} âˆ›y = log_{10} z

log_{10} **âˆš**x/âˆ›y = log_{10} z

**âˆš**x/âˆ›y = z

z = **âˆš**x/âˆ›y

**15.** **Given that log m = x + y and log n = x – y, express the value of log mÂ²n in terms of x and y. **

**Solution:**

Given:

log m = x + y

log n = x â€“ y

log mÂ²n

Let us simplify the given expression,

log mÂ²n = log m^{2} + log n

= 2 log m + log n

By substituting the given values, we get

= 2 (x + y) + (x – y)

= 2x + 2y + x â€“ y

= 3x + y

**16. Given that log x = m + n and log y = m â€“ n, express the value of log (10x/y ^{2}) in terms of m and n.**

**Solution:**

Given:

log x = m + n

log y = m â€“ n

log (10x/y^{2})

Let us simplify the given expression,

log (10x/y^{2}) = log 10x â€“ log y^{2}

= log 10 + log x â€“ 2 log y

= 1 + log x â€“ 2 log y

= 1 + (m + n) â€“ 2(m – n)

= 1 + m + n â€“ 2m + 2n

= 1 â€“ m + 3n

**17. If log x/2 = log y/3, find the value of y ^{4}/x^{6}.**

**Solution:**

Given:

log x/2 = log y/3

Let us simplify the given expression,

By cross multiplying, we get

3 log x = 2 log y

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

so, x^{3} = y^{2}

now square on both sides, we get

(x^{3})^{2} = (y^{2})^{2}

x^{6} = y^{4}

y^{4}/x^{6} = 1

**18. Solve for x:**

**(i) log x + log 5 = 2 log 3**

**(ii) log _{3} x â€“ log_{3} 2 = 1**

**(iii) x = log 125/log 25**

**(iv) (log 8/log 2) Ã— (log 3/logâˆš3) = 2 log x**

**Solution:**

**(i) **log x + log 5 = 2 log 3

Let us solve for x,

Log x = 2 log 3 â€“ log 5

= log 3^{2} â€“ log 5

= log 9 â€“ log 5

= log (9/5)

âˆ´ x = 9/5

**(ii) **log_{3} x â€“ log_{3} 2 = 1

Let us solve for x,

log_{3} x = 1 + log_{3} 2

= log_{3} 3 + log_{3} 2 [Since, 1 can be written as log_{3} 3 = 1]

= log_{3} (3Ã—2)

= log_{3} 6

âˆ´ x = 6

**(iii) **x = log 125/log 25

x = log 5^{3}/log5^{2}

** = **3 log 5/ 2 log 5

= 3/2 [Since, log 5/log 5 = 1]

âˆ´ x = 3/2

**(iv) **(log 8/log 2) Ã— (log 3/logâˆš3) = 2 log x

(log 2^{3}/log 2) Ã— (log 3/log3^{1/2}) = 2 log x

(3log 2/log 2) Ã— (log 3/Â½ log 3) = 2 log x

3 Ã— 1/(Â½) = 2 log x

3 Ã— 2 = 2 log x

6 = 2 log x

log x = 6/2

log x = 3

x = (10)^{3}

= 1000

âˆ´ x = 1000

**19.** **Given 2 log _{10 }x + 1= log_{10 }250, find **

**(i) x**

**(ii) log**

Solution:

_{10}2xSolution:

Given:

2 log_{10 }x + 1= log_{10 }250

(i) let us simplify the above expression,

log_{10} x^{2} + log_{10} 10 = log_{10 }250 [Since, 1 can be written as log_{10} 10]

log_{10} (x^{2} Ã— 10) = log_{10 }250

(x^{2} Ã— 10) = 250

x^{2} = 250/10

x^{2} = 25

x = âˆš25

= 5

âˆ´ x = 5

**(ii) **log_{10 }2x

We know that, x = 5

So, log_{10 }2x = log_{10} 2Ã—5

= log_{10} 10

= 1

**20. If log x/log 5 = log y ^{2}/log 2 = log 9/log (1/3), find x and y.**

**Solution:**

Given:

log x/log 5 = log y^{2}/log 2 = log 9/log (1/3)

let us consider,

log x/log 5 = log 9/log (1/3)

log x = (log 9Ã—log 5)/log (1/3)

= (log 3^{2} Ã— log 5) / (log 1 â€“ log 3)

= (2 log 3 Ã— log 5) / (-log 3) [log 1 = 0]

= -2 Ã— log 5

= log 5^{-2}

x = 5^{-2}

= 1/5^{2}

= 1/25

Now,

log y^{2}/log 2 = log 9/log (1/3)

log y^{2} = (log 9Ã—log2)/log (1/3)

= (log 3^{2} Ã— log 2) / (log 1 â€“ log 3)

= (2 log 3 Ã— log 2) / (-log 3) [log 1 = 0]

= -2 Ã— log 2

= log 2^{-2}

y^{2} = 2^{-2}

= 1/2^{2}

= Â¼

= âˆš(1/4)

= Â½

**21. Prove the following: (i) 3 ^{log 4}Â = 4^{log 3}(ii) 27^{log 2}Â = 8^{log 3}Solution:**

**(i) **3^{log 4}Â = 4^{log 3}

Let us take log on both sides,

If log 3^{log 4}Â = log 4^{log 3}

log 4 . log 3 = log 3 . log 4

log 2^{2} . log 3 = log 3 . log 2^{2}

2 log 2 . log 3 = log 3 . 2 log 2

Which is true.

Hence proved.

**(ii) **27^{log 2}Â = 8^{log 3}

Let us take log on both sides,

If log 27^{log 2}Â = log 8^{log 3}

log 2 . log 27 = log 3 . log 8

log 2 . log 3^{3} = log 3 . log 2^{3}

log 2 . 3 log 3 = log 3 . 3 log 2

3 log2 . log 3 = 3 log2 . log 3

Which is true.

Hence proved.

**22.** **Solve the following equations:**

**(i) log (2x + 3) = log 7**

**(ii) log (x +1) + log (x â€“ 1) = log 24**

**(iii) log (10x + 5) â€“ log (x â€“ 4) = 2**

**(iv) log _{10 }5 + log_{10 }(5x + 1) = log_{10 }(x + 5) + 1**

**(v) log (4y â€“ 3) = log (2y + 1) â€“ log 3**

**(vi) log**

_{10 }(x + 2) + log_{10 }(x â€“ 2) = log_{10}3 + 3 log_{10 }4**(vii) log (3x + 2) + log (3x â€“ 2) = 5 log 2**

**Solution:**

**(i) **log (2x + 3) = log 7

Let us simplify the expression,

2x + 3 = 7

2x = 7 â€“ 3

2x = 4

x = 4/2

= 2

**(ii) **log (x +1) + log (x â€“ 1) = log 24

Let us simplify the expression,

log [(x +1) (x â€“ 1)] = log 24

log (x^{2} – 1) = log 24

(x^{2} – 1) = 24

x^{2} = 24 + 1

= 25

x = âˆš25

= 5

**(iii) **log (10x + 5) â€“ log (x â€“ 4) = 2

Let us simplify the expression,

log (10x + 5) / (x â€“ 4) = 2 log 10

log (10x + 5) / (x â€“ 4) = log 10^{2}

(10x + 5) / (x â€“ 4) = 100

10x + 5 = 100 (x – 4)

10x + 5 = 100x â€“ 400

5 + 400 = 100x â€“ 10x

90x = 405

x = 405/90

= 81/18

= 9/2

= 4.5

**(iv) **log_{10 }5 + log_{10 }(5x + 1) = log_{10 }(x + 5) + 1

Let us simplify the expression,

log_{10 }[5Ã— (5x + 1)] = log_{10 }(x + 5) + log_{10} 10

log_{10 }[5Ã— (5x + 1)] = log_{10} [(x + 5) Ã— 10]
[5Ã— (5x + 1)] = [(x + 5) Ã— 10]

25x + 5 = 10x + 50

25x â€“ 10x = 50 â€“ 5

15x = 45

x = 45/15

= 3

**(v) **log (4y â€“ 3) = log (2y + 1) â€“ log 3

Let us simplify the expression,

log (4y â€“ 3) = log (2y + 1) / 3

(4y â€“ 3) = (2y + 1) / 3

By cross multiplying, we get

3(4y – 3) = 2y + 1

12y â€“ 9 = 2y + 1

12y â€“ 2y = 9 + 1

10y = 10

y = 10/10

= 1

**(vi) **log_{10 }(x + 2) + log_{10 }(x â€“ 2) = log_{10}3 + 3 log_{10 }4

Let us simplify the expression,

log_{10} [(x + 2) Ã— (x – 2)] = log_{10} 3 + log_{10} 4^{3}

log_{10} [(x + 2) Ã— (x – 2)] = log_{10} (3Ã—4^{3})

^{3})

(x^{2} – 4) = (3Ã—4Ã—4Ã—4)

(x^{2} – 4) = 192

x^{2} = 192 + 4

= 196

x = âˆš196

= 14

**(vii) **log (3x + 2) + log (3x â€“ 2) = 5 log 2

Let us simplify the expression,

log (3x + 2) + log (3x â€“ 2) = log 2^{5}

log [(3x + 2) Ã— (3x – 2)] = log 32

log (9x^{2} – 4) = log 32

(9x^{2} – 4) = 32

9x^{2} = 32 + 4

9x^{2} = 36

x^{2} = 36/9

x^{2} = 4

x = âˆš4

= 2

**23. Solve for x: log _{3}Â (x + 1) â€“ 1 = 3 + log_{3}Â (x â€“ 1)Solution:**

Given:

log_{3}Â (x + 1) â€“ 1 = 3 + log_{3}Â (x â€“ 1)

Let us simplify the expression,

log_{3}Â (x + 1) â€“ log_{3}Â (x â€“ 1) = 3 + 1

log_{3}Â (x + 1) / (x â€“ 1) = 4 log_{3} 3 [Since, log_{3} 3 = 1]

log_{3}Â (x + 1) / (x â€“ 1) = log_{3} 3^{4}

(x + 1) / (x â€“ 1) = 3^{4}

By cross multiplying, we get

(x + 1) = 81 (x – 1)

x + 1 = 81x â€“ 81

81x â€“ x = 1 + 81

80x = 82

x = 82/80

= 41/40

= 1 1/40

**24. Solve for x:**

**5 ^{log x} + 3^{log x} = 3^{log x+1} â€“ 5^{log x â€“ 1}**

**Solution:**

Given:

5^{log x} + 3^{log x} = 3^{log x+1} â€“ 5^{log x â€“ 1 }

Let us simplify the expression,

5^{log x} + 3^{log x} = 3^{log x} . 3^{1} â€“ 5^{log x} . 5^{-1}

5^{log x} + 3^{log x} = 3.3^{log x} â€“ 1/5 . 5^{log x}

5^{log x} + 1/5 . 5^{log x} = 3.3^{log x} â€“ 3^{log x}

(1 + 1/5) 5^{log x} = (3 – 1) 3^{log x}

(6/5) 5^{log x} = 2(3^{log x})

5^{log x} / 3^{log x} = (2Ã—5)/6

(5/3)^{log x} = 10/6

(5/3)^{log x} = 5/3

(5/3)^{log x} = (5/3)^{1}

So, by comparing the powers

log x = 1

log x = log 10

x = 10

**25. If log (x-y)/2 = Â½ (log x + log y), prove that x ^{2} + y^{2} = 6xy**

**Solution:**

Given:

log (x-y)/2 = Â½ (log x + log y)

Let us simplify,

log (x-y)/2 = Â½ (log xÃ—y)

log (x-y)/2 = Â½ log xy

log (x-y)/2 = log (xy)^{1/2 }

(x-y)/2 = (xy)^{1/2}

By squaring on both sides, we get

[(x-y)/2]^{2}= [(xy)

^{1/2}]

^{2}

(x – y)^{2}/4 = xy

By cross multiplying, we get

(x – y)^{2} = 4xy

x^{2} + y^{2} â€“ 2xy = 4xy

x^{2} + y^{2} = 4xy + 2xy

x^{2} + y^{2} = 6xy

Hence proved.

**26. If x ^{2} + y^{2} = 23xy, Prove that log (x + y)/5 = Â½ (log x + log y)**

**Solution:**

Given:

x^{2} + y^{2} = 23xy

So, the above equation can be written as

x^{2} + y^{2} = 25xy â€“ 2xy

x^{2} + y^{2} + 2xy = 25xy

(x + y)^{2} = 25xy

(x + y)^{2} / 25 = xy

Now by taking log on both sides, we get

log [(x + y)^{2} / 25] = log xy

log [(x + y)/5]^{2} = log xy

2 log (x+y)/5 = log x + log y

log (x+y)/5 = Â½ log x + log y

Hence proved.

**27. If p = log _{10} 20 and q = log_{10} 25, find the value of x if 2 log_{10} (x + 1) = 2p â€“ q**

**Solution:**

Given:

p = log_{10} 20

q = log_{10} 25

Then,

2 log_{10} (x + 1) = 2p â€“ q

Now substitute the values of p and q, we get

2 log_{10} (x + 1) = 2 log_{10} 20 – log_{10} 25

= 2 log_{10} 20 – log_{10} 5^{2}

= 2 log_{10} 20 â€“ 2 log_{10} 5

2 log_{10} (x + 1) = 2 (log_{10} 20 â€“ log_{10} 5)

log_{10} (x + 1) = (log_{10} 20 â€“ log_{10} 5)

= log_{10} (20/5)

log_{10} (x + 1) = log_{10} 4

(x + 1) = 4

x = 4 – 1

= 3

**28. Show that:**

**(i) 1/log _{2} 42 + 1/log_{3} 42 + 1/log_{7} 42 = 1**

**(ii) 1/log _{8} 36 + 1/log_{9} 36 + 1/log_{18} 36 = 2**

**Solution:**

**(i) **1/log_{2} 42 + 1/log_{3} 42 + 1/log_{7} 42 = 1

Let us consider LHS:

1/log_{2} 42 + 1/log_{3} 42 + 1/log_{7} 42

By using the formula, log_{n }m = log_{m} / log_{n}

1/log_{2} 42 + 1/log_{3} 42 + 1/log_{7} 42 = 1/(log 42/log_{2}) + 1/(log 42/log_{3}) + 1/(log 42/log_{7})

= log_{2}/log 42 + log_{3}/log 42 + log_{7}/log 42

= (log_{2} + log_{3} + log_{7})/log 42

= (log 2Ã—3Ã—7)/log 42

= log 42 / log 42

= log 42/log 42

= 1

= RHS

**(ii) **1/log_{8} 36 + 1/log_{9} 36 + 1/log_{18} 36 = 2

Let us consider LHS:

1/log_{8} 36 + 1/log_{9} 36 + 1/log_{18} 36

By using the formula, log_{n }m = log_{m} / log_{n}

1/log_{8} 36 + 1/log_{9} 36 + 1/log_{18} 36 = 1/(log 36/log_{8}) + 1/(log 36/log_{9}) + 1/(log 36/log_{18})

= log_{8}/log 36 + log_{9}/log 36 + log_{18}/log 36

= (log_{8} + log_{9} + log_{18})/log 36

= (log 8Ã—9Ã—18)/log 36

= log 36^{2}/log 36

= 2 log 36/log 36

= 2

= RHS

**29. Prove the following identities:**

**(i) 1/log _{a} abc + 1/log_{b} abc + 1/log_{c} abc = 1**

**(ii) log _{b} a. log_{c} b. log_{d} c = log_{d} a**

**Solution:**

**(i) **1/log_{a} abc + 1/log_{b} abc + 1/log_{c} abc = 1

Let us consider LHS:

1/log_{a} abc + 1/log_{b} abc + 1/log_{c} abc

By using the formula, log_{n }m = log_{m} / log_{n}

1/log_{a} abc + 1/log_{b} abc + 1/log_{c} abc = 1/(log abc/log_{a}) + 1/(log abc/log_{b}) + 1/(log abc/log_{c})

= log_{a}/ log abc + log_{b}/ log abc + log_{c}/ log abc

= (log_{a} + log_{b} + log_{c})/ log abc

= (log aÃ—bÃ—c)/ log abc

= log abc/log abc

= 1

= RHS

**(ii) **log_{b} a. log_{c} b. log_{d} c = log_{d} a

Let us consider LHS:

log_{b} a. log_{c} b. log_{d} c = (log a/ log b) Ã— (log b/ log c) Ã— (log c/ log d)

= log a/log d

= log_{d} a

= RHS

**30. Given that log _{a} x = 1/ Î±, log_{b} x = 1/Î², log_{c} x = 1/Î³, find log_{abc} x.**

**Solution:**

It is given that:

log_{a} x = 1/ Î±, log_{b} x = 1/Î², log_{c} x = 1/Î³

So,

log_{a} x = 1/ Î± => log x/log_{a} _{ }= 1/ Î± => log_{a} = Î± log x

log_{b} x = 1/Î² => log x/log_{b} _{ }= 1/ Î² => log_{b} = Î² log x

log_{c} x = 1/Î³ => log x/log_{c} _{ }= 1/ Î³ => log_{c} = Î³ log x

Now,

log_{abc }x = log x/log abc

= log x/(log a + log b + log c)

= log x/(Î± log x + Î² log x + Î³ log x)

= log x/log x(Î±+ Î²+ Î³)

= 1/(Î±+ Î²+ Î³)

**31. Solve for x:**

**(i) log _{3} x + log_{9} x + log_{81} x = 7/4**

**(ii) log _{2} x + log_{8} x + log_{32} x = 23/15**

**Solution:**

**(i) **log_{3} x + log_{9} x + log_{81} x = 7/4

let us simplify the expression,

1/log_{x} 3 + 1/log_{x} 9 + 1/log_{x} 81 = 7/4

1/log_{x} 3^{1} + 1/log_{x} 3^{2} + 1/log_{x} 3^{4} = 7/4

1/log_{x} 3 + 1/2log_{x} 3 + 1/4log_{x} 3 = 7/4

1/log_{x} 3 [1 + Â½ + Â¼] = 7/4

1/log_{x} 3 [(4+2+1)/4] = 7/4

log_{3} x [7/4] = 7/4

log_{3} x = (7/4) Ã— (4/7)

log_{3} x = 1

log_{3} x = log_{3} 3 [Since, 1= log_{a }a]

On comparing, we get

x = 3

**(ii) **log_{2} x + log_{8} x + log_{32} x = 23/15

let us simplify the expression,

1/log_{x} 2 + 1/log_{x} 8 + 1/log_{x} 32 = 23/15

1/log_{x} 2^{1} + 1/log_{x} 2^{3} + 1/log_{x} 2^{5} = 23/15

1/log_{x} 2 + 1/3log_{x} 2 + 1/5log_{x} 2 = 23/15

1/log_{x} 2 [1 + 1/3 + 1/5] = 23/15

log_{2} x [(15 + 5 + 3)/15] = 23/15

log_{2} x [23/15] = 23/15

log_{2} x = (23/15) Ã— (15/23)

log_{2} x = 1

log_{2} x = log_{2} 2 [Since, 1= log_{a }a]

On comparing, we get

x = 2