Question

please solve all the problems and get brainest​

Answer 1

Question 32 :

Given that log 2 = 0.3010 and log 3 = 0.4771, find the value of the following :

Answer :

(i) log 12

= log ( 3 × 4 )

= log ( 3 × 2² )

= log 3 + log 2²

= log 3 + 2log 2

= 0.4771 + 2( 0.3010 )

= 0.4771 + 0.6020

= 1.0791

(ii) log 5

= log 10/2

= log 10 - log 2

= 1 - log 2

= 1 - 0.3010

= 0.699

(iii) log 5^( 1/3 )

= 1/3 × log 5

= 1/3 × 0.699

= 0.233

(iv) log 108

= log ( 27 × 4 )

= log ( 3³ × 2² )

= log 3³ + log 2²

= 3log 3 + 2log 2

= 3( 0.4771 ) + 2( 0.3010 )

= 1.4313 + 0.6020

= 2.0333

(v) log (3/8)

= log ( 3 / 2³ )

= log 3 - log 2³

= log 3 - 3log 2

= 0.4771 - 3( 0.3010 )

= 0.4771 - 0.903

= - 0.4259

(vi) log 48

= log ( 16 × 3 )

= log ( 2^4 × 3 )

= log 2^4 + log 3

= 4log 2 + log 3

= 4( 0.3010 ) + 0.4771

= 1.204 + 0.4771

= 1.6811

Question 33 :

Evaluate each of the following :

Answer :

(i) 3^[ - (1/2) log_3 81 ]

= 3^[ - (1/2) log_3 3^4 ]

= 3^[ - (1/2) 4 × log_3 3 ]

= 3^[ - (1/2) × 4 ]

= 3^[ - 2 ]

= 1 / 3²

= 1 / 9

(ii) 2^[ log_(2√2) 15 ]

$\sf = 2^{log_{ 2^{3/2 } } 15 }$

Since log_( a^m ) n = 1/m × log_a n

$\sf = 2^{ \frac{2}{3} \times log_2 15 }$

Using a^( mn)  = [ a^m ]^n

$\sf = (2^{ log_2 15 })^{ \frac{2}{3} }$

Since a^( log_a N) = N

= 15^( 2/3 ) ≈ 6.0822

$\sf (iv)\ \ {a}^{ log_{a}m + log_{a}n+ log_{a}p }$

$\sf = {a}^{ log_{a}mnp }$

Since a^( log_a N) = N

= mnp

(v)  log_π ( tan 0.25π )

= log_π ( tan π/4 )

= log_π 1

= 0

(vi) log_2 ( log_3 81 )

= log_2 ( log_3 3^4 )

= log_2 ( 4log_3 3 )

= log_2 4

= log_2 2²

= 2log_2 2

= 2

(vii)  log_3 5 × log_25 27

= ( log 5 / log 3 ) × ( log 27 / log 25 )

= ( log 5 / log 3 ) × ( log 3³ / log 5² )

= ( log 5 / log 3 ) × ( 3log 3 / 2log 2)

= 3/2

Question 34 :

Prove that :

(i) a^[ log_a 1 + 2log_a 2 + 3log_a 3 + ........ + nlog_a n = 2² × 3³ × 4^4 × ........ n^n

Answer :

$\sf a^{log_a 1 + 2log_a 2 + 3log_a 3 + ........ + nlog_a n }= {2}^{2} \times {3}^{3} \times {4}^{4} ........ {n}^{n}$

Consider LHS

$\sf = a^{log_a 1 + 2log_a 2 + 3log_a 3 + ........ + nlog_a n }$

Using Power rule m. log a = log a^m

$\sf = a^{log_a 1 + log_a 2^{2} + log_a 3^{3} + ........ + log_a n ^{n} }$

Using Product rule log a + log b = log ab

$\sf = a^{log_a (1 \times 2^{2} \times 3^{3} \times ........ \times n ^{n} )}$

$\sf = a^{log_a (2^{2} \times 3^{3} \times ........ \times n ^{n} )}$

$\sf = 2^{2} \times 3^{3} \times ........ \times n ^{n}$

= RHS

Hence proved.

Question 34 :

Prove that :

(ii)  a^[ log_a 1 + 2log_a 2 + 2log_a 3 + ...... + 2.log_a n ] = ( n! )²

Answer :

$\sf a^{log_a 1 + 2log_a 2 + 2log_a 3 + ...... + 2.log_a n } = ( n! )^{2}$

Consider LHS

$\sf = a^{log_a 1 + 2log_a 2 + 2log_a 3 + ...... + 2.log_a n }$

$\sf = a^{log_a 1 + log_a {2}^{2} + log_a {3}^{2} + ...... + log_a {n}^{2} }$

$\sf = a^{log_a ( {2}^{2} \times {3}^{2} \times ...... \times {n}^{2} )}$

$\sf = a^{log_a ({ 2 \times 3 \times ...... \times n)}^{2} }$

$\sf = a^{log_a ({n!)}^{2} }$

$\sf = ({n!)}^{2}$

= RHS

Hence proved.

Question 35 :

Prove that :

(i) log_(10) tan 1° × log_(10) tan 2° × log_(10) tan 3° × ........ log_(10) tan 50° = 0

Answer :

$\sf log_{10} tan 1 \times log_{10} tan 2 \times log_{10} tan 3 \times ........ log_{10} tan 50= 0$

Consider LHS

$\sf = log_{10} tan 1 \times log_{10} tan 2 \times log_{10} tan 3 \times ........ log_{10} tan 50$

It can be written as

$\sf = log_{10} tan 1 \times log_{10} tan 2 \times log_{10} tan 3 \times ........ \times log_{10}tan45 \times ....... \times log_{10} tan 50$

Substituting tan 45° = 1

$\sf = log_{10} tan 1 \times log_{10} tan 2 \times log_{10} tan 3 \times ........ \times log_{10}1 \times ....... \times log_{10} tan 50$

Since log_a 1 = 0

$\sf = log_{10} tan 1 \times log_{10} tan 2 \times log_{10} tan 3 \times ........ \times 0 \times ....... \times log_{10} tan 50$

= 0

= RHS

Hence proved.

Question 35:

Prove that :

(ii) log_(10) tan 1° + log_(10) tan 2° + log_(10) tan 3° + ........ log_(10) tan 89° = 0

Answer :

Consider LHS

= log_(10) tan 1° + log_(10) tan 2° + ........ log_(10) tan 88° + log_(10) tan 89°

= log_(10) [ tan 1° × tan 2° × ....... tan 88° × tan 89° ]

It can be written as

= log_(10) [ tan 1° × tan 2° × ....... tan ( 90 - 2 )° × tan ( 90 - 1 )° ]

Using trigonometric ratios for complementary angles tan ( 90 - Φ ) = cot Φ

= log_(10) [ ( tan 1° × cot 1° ) × ( tan 2° × cot 2° ) ..... ]

= log_(10) [ ( tan 1° × 1 / tan 1° ) × ( tan 2° × 1 / tan 2° ) ....]

= log_(10) [ 1 × 1 × ....... 1 ]

= log_(10) 1

= 0

= RHS

Hence proved

Answer 2

Refer to the attachment for qn 32

Question 33 :

Answer :

(i) 3^[ - (1/2) log_3 81 ]

= 3^[ - (1/2) log_3 3^4 ]

= 3^[ - (1/2) 4 × log_3 3 ]

= 3^[ - (1/2) × 4 ]

= 3^[ - 2 ]

= 1 / 3²

= 1 / 9

(ii) 2^[ log_(2√2) 15 ]

Since log_( a^m ) n = 1/m × log_a n

Using a^( mn)  = [ a^m ]^n

Since a^( log_a N) = N

= 15^( 2/3 ) ≈ 6.0822

Since a^( log_a N) = N

= mnp

(v)  log_π ( tan 0.25π )

= log_π ( tan π/4 )

= log_π 1

= 0

(vi) log_2 ( log_3 81 )

= log_2 ( log_3 3^4 )

= log_2 ( 4log_3 3 )

= log_2 4

= log_2 2²

= 2log_2 2

= 2

(vii)  log_3 5 × log_25 27

= ( log 5 / log 3 ) × ( log 27 / log 25 )

= ( log 5 / log 3 ) × ( log 3³ / log 5² )

= ( log 5 / log 3 ) × ( 3log 3 / 2log 2)

= 3/2

Question 34 :

Prove that :

(i) a^[ log_a 1 + 2log_a 2 + 3log_a 3 + + nlog_a n = 2² × 3³ × 4^4 × n^n

Answer :

Consider LHS

Using Power rule m. log a = log a^m

Using Product rule log a + log b = log ab

= RHS

Hence proved.

Question 34 :

Prove that :

(ii)  a^[ log_a 1 + 2log_a 2 + 2log_a 3 + + 2.log_a n ] = ( n! )²

Answer :

Consider LHS= RHS

Hence proved.

Question 35 :

Prove that :

(i) log_(10) tan 1° × log_(10) tan 2° × log_(10) tan 3° × log_(10) tan 50° = 0

Answer :

Consider LHS

It can be written as

Substituting tan 45° = 1

Since log_a 1 = 0= 0 = RHS

Hence proved.

Question 35:

Prove that :

(ii) log_(10) tan 1° + log_(10) tan 2° + log_(10) tan 3° + log_(10) tan 89° = 0

Answer :

Consider LHS

= log_(10) tan 1° + log_(10) tan 2° + log_(10) tan 88° + log_(10) tan 89°

= log_(10) [ tan 1° × tan 2° ×tan 88° × tan 89° ]

It can be written as

= log_(10) [ tan 1° × tan 2° × tan ( 90 - 2 )° × tan ( 90 - 1 )° ]

Using trigonometric ratios for complementary angles tan ( 90 - Φ ) = cot Φ

= log_(10) [ ( tan 1° × cot 1° ) × ( tan 2° × cot 2° ) . ]

= log_(10) [ ( tan 1° × 1 / tan 1° ) × ( tan 2° × 1 / tan 2° ) ]

= log_(10) [ 1 × 1 × 1 ]

= log_(10) 1

= 0 = RHS