## Rajasthan Board RBSE Class 11 Maths Chapter 9 Logarithms Ex 9.2

Question 1.

Prove that:

log 630 = log 2 + 2 log 3 + log 5 + log 7.

Solution:

L.H.S. = log 630

= log (2 x 3 x 3 x 5 x 7)

= log (2 x 32 x 5 x 7)

= log 2 + log 3^{2} + log 5 + log 7

= log 2 + 2 log 3 + log 5 + log 7

Hence Proved.

Question 2.

Prove that:

Solution:

Question 3.

Prove that:

log 10 + log 100 + log 1000 + log 10000 = 10

Solution:

L.H.S.

= log 10 + log 10^{2} + log 10^{3} + log 10^{4}

= log 10 + 2 log 10 + 3 log 10 + 4 log 10

= 10 log 10

= 10 x 1 (∵ log 10 = 1)

= 10

= R.H.S.

Hence Proved.

Question 4.

If log 2 = 0.3010, log 3 = 0.4771, log 7 = 0.8451 and log 11 = 1.0414, then find the value of the following :

Solution:

(i) log 36 = log (2 x 2 x 3 x 3)

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

= log 2^{2} + log 3^{2}

= 2 log 2 + 2 log 3

= 2(log 2 + log 3)

= 2(0.3010 + 0.4771)

= 2(0.7781)

= 1.5562

= log(2 x 3 x 7) – log 11

= log 2 + log 3 + log 7 – log 11

= 0.3010 + 0.4771 + 0.8451 – 1.0414

= 1.6232 – 1.0414

= 0.5818

= 5(log 11 – log 7)

= 5(1.0414 – 0.8451)

= 5(0.1963)

= 0.9815

(iv) log 70 = log (7 x 10)

= log 7 + log 10

= 0.8451 + 1

= 1.8451

= log 11^{2} – log( 12 x 10)

= 2 log 11 – log 12 – log 10

= 2(1.0414)- log (2 x 2 x 3) – 1

= 2.0828 – 1 – 2 log 2 – log 3

= 1.0828 – 2(0.3010) – 0.4771

= 1.0828 – 0.6020 – 0.4771

= 1.0828 – 1.0791

= 0.0037

Question 5.

Find the value of x from following equation

log_{x}4 + log_{x}16 + log_{x}64 = 12

Solution :

log_{x} 4 + log_{x} 16 + log_{x} 64 = 12

⇒ log_{x} 4 + log_{x} 42 + log, 43 = 12

⇒ log_{x} 4 + 2 log_{x} 4 + 3 log_{x} 4=12

⇒ 6 log_{x} 4 = 12

⇒ log_{x} 4 = 2

⇒ log_{x} 2^{x} = 2

⇒ 2 log_{x }2 = 2

⇒ log_{x }2 = 1

⇒ log_{x }2 = log_{2 }2

On comparing, x = 2

Question 6.

Solve the equation :

log (x + 1) – log (x – 1) = 1

Solution:

log (x + 1) – log (x – 1) = 1

Question 7.

Find the value of 3^{2}^{-log34}

Solution:

3^{2} – log_{3}^{4} = 3^{2} log_{3}3 – log^{3}4

= 3log_{3} 3^{2} – log_{3}^{4}

= 3log_{3} 9 – log_{3} 4

= 3log_{3}(9/4)

If log M = – 2, 1423 then to get its characteristic and mantissa is made positive as following:

= [latex]\frac { 9 }{ 4 }[/latex] = 2 [latex]\frac { 1 }{ 4 }[/latex] (∵ a^{logax} = x)

Question 8.

Give the solution of following questions in one term :

(i) log 2 + 1

(ii) log 2x + 2 log x

Solution : (i) log 2 + 1

= log 2 + log 10 (∵ log_{10}10 = 1)

= log (2 x 10)

= log 20

(ii) log 2x + 2 log x

= log 2x + log (x)^{2}

= log (2x × x^{2})

= log2x^{3}

Question 9.

Prove that:

(i) log_{5}3 . log_{3}4 . log_{2}5 = 2

(ii) log_{a}x × log_{b}y = log_{b}x × log_{a} y.

Solution:

(i) L.H.S. = log_{5}3 . log_{3}4. log_{2}5