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(i) Number 1270 is 4 digit number.

So, characteristic of its logarithm will be 4 – 1 = 3.

(ii) In 20.125, integral part is 20 which contains 2 digit.

So, characteristic of its logarithm will be 2 – 1 = 1.

(iii) In 7.985, integral part is 7 which contains 1 digit.

So, characteristic of its logarithm will be 1 – 1 = 0.

(iv) In 431.5, integral part is 431 which contains 3 digit.

So, characteristic of its logarithm will be 3 – 1 = 2.

(v) In 0.02, there are 1 zero between decimal point and first significant digit.

So, characteristic of its logarithm will be – (1 + 1) = – 2 or \(\bar{2}\) .

(vi) In 0.02539, there are zero between decimal point and first significant digit.

So, characteristic of its logarithm will be – (1 + 1) = – 2 or \(\bar{2}\).

(vii) Number 70 is 2 digit number.

So, characteristic of its logarithm will be 2 – 1 = 1.

(viii) In 0.000287, there are 3 zero between decimal point and first significant digit.

So, characteristic of its logarithm will be -(3 + 1) = -4 or \(\bar{4}\).

(ix) In 0.005, there are 2 zero between decimal point and first significant digit.

So, characteristic of its logarithm will be – (2 + 1) = – 3 or \(\bar{3}\).

(x) In 0.00003208, there are 4 zero between decimal point and first significant digit.

So, characteristic of its logarithm will be – (4 + 1) = – 5 or \(\bar{5}\) .

(xi) In 0.000485, there are 3 zero between decimal point and first significant digit.

So, characteristic of its logarithm will be – (3 + 1) = -4 or \(\bar{4}\).

(xii) In 0.007, there are 2 zero between decimal point and first significant digit.

So, characteristic of its logarithm will be – (2 + 1) = – 3 or \(\bar{3}\).

(xiii) In 0.0005309, there are 3 zero between decimal point and first significant digit.

So, characteristic of its logarithm will be – (3 + 1 ) = – 4 or \(\bar{4}\).

(i) 5^-2 = 1/25

Logarithm form of 5^-2 = 1/25 is

log₅(1/25) = -2.

(ii) 10^-3 = 0.001

Logarithm form of 10^-3 = 0.001 is

log₁₀0.001 = – 3

(iii) 4^3/2 = 8

Logarithm form of 4^3/2 = 8 is

log₄ 8 = 3/2

(i) 2⁶ = 64

Logarithm form of 2⁶ = 64 is

log₂ 64 = 6

(ii) 10⁴ = 10000

Logarithm form of 10⁴ = 10000 is

log₁₀10000 = 4

(iii) 2¹⁰ = 1024

Logarithm form of 2¹⁰= 1024 is

log₂ 1024 = 10

(a) Given, antilog 1.5662 = 36.83

(i) ∵ Mantissa of 1.5662 and 1.5662 are same. 

So, anitlog of two number will be same digit numbers.

Now, characteristics of which number is 1 then there will be no zero after decimal point in antilog.

∴ Required antilog \(\bar{1}\).5662 = 0.3683.

(ii) ∵ Mantissas of 1.5662 and 2.5662 are same.

So, antilog of two numbers will be same digit number.

Now, characteristic of which number is 2 then decimal point will be after 3 digit in antilog.

∴ Required anitlog 2.5662 = 368.3.

(iii) ∵ Mantìssa of 1.5662 and \(\bar{2}\).5662 are same.

So. antilog of two numbers will be same digit number.

Now, characteristic of which number is \(\bar{2}\) then there will be one zero after decimal point in anitlog.

∴ Required antilog \(\bar{2}\).5662 = 0.03683.

(b) antilog and log are opposite to each other.

∴ antilog (log x) x.

(i) 2813

Characteristic : There are characteristic will be 4 – 1 = 3.

Mantissa : In log table, in first column, in front of 28 and under the column 1.

Find the number which is = 4487.

For fourth digit 3’s mean difference = 5

On adding = 4492

So, mantissa of log₁₀ 2813 = 0.4492

Thus, log₁₀ 2813 = Characteristic + Mantissa

= 3 + 0.4492 = 3.4492

(ii) 400

Characteristic : There are 3 digit in number 400.

So, characteristic will be 3 – 1 = 2.

Mantissa : In log table, in first column, in front of 40 and under the column 0, find the number which is 6021.

So, mantissa of log₁₀ 400 = 0.6021

Thus, log₁₀ 400 = Characteristic + Mantissa

= 2 + 0.6021 =2.6021

(iii) 27.28

Characteristic : 

Here, integral part is 27 which contains 2 digit. 

So, characteristic of its logarithm will be 2 – 1 = 1.

Mantissa : In log table, in first column, in front of 27 (ignoring decimal point) and under the column 2, find the number which is = 4346

For fourth digit 8’s mean difference = 13

On adding = 4359

So, mantissa of log₁₀ 27.28 = 0.4359

Thus, log₁₀ 27.28 = Characteristic + Mantissa

= 1+ 0.04359 =1.4359

(iv) 9

Characteristic : There are only 1 digit in number 9.

So, characteristic will be 1 – 1 = 0.

Mantissa : In log table, in first column, in front of 90 and under the column 0 find the number which is = 9542.

So mantissa of log₁₀ 9 = 0.9542

Thus, log₁₀ 9 = Characteristic + Mantissa

= 0 + 0.9542 = 0.9542

(v) 0-678

Characteristic : In 0.678, there are no zero between decimal point and first significant digit. 

So, characteristic of its logarithm will be – (0+ 1) = – 1 = 1

Mantissa : In log table, in first column, in front of 67 (ignoring decimal point) and under the column 8 find the number which is 8312.

So, mantissa of log₁₀ 0.678 = 0.8312
Thus,
log₁₀ 0.678 = Characteristic + Mantissa

= \(\bar{1}\) + 0.8312 = \(\bar{1}\). 8312

(vi) 0.0035

Characteristic : In 0.0035, there are 2 zero between decimal point and first significant digit. 

So, characteristic of its logarithm will be
-(2+1) = -3 = \(\bar{3}\)

Mantissa : In log table, in first column, in front of 35 (ignoring decimal point) and under the column 0, find the number which is = 5441

So, mantissa of log₁₀0.0035 = 0.5441

Thus, log₁₀ 0.0035 = Characteristic + Mantissa

= \(\bar{3}\) + 0.5441 = \(\bar{3}\).5441

(vii) 0.08403

Characteristic : In 0.08403, there are 1 zero between decimal point and first significant digit. 

So, characteristic of its logarithm will be
-(1 + 1) = – 2 = 2

Mantissa : In log table, in first column, in front of 84 (ignoring decimal point) and under the column 0 find the number, which is = 9243

For fourth digit 3 mean difference = 2

On adding = 9245

So, mantissa of log₁₀0.08403 = 0.9245

Thus,
log₁₀ 0.08403 = Characteristic + Mantissa

= \(\bar{2}\) + 0.9245 = \(\bar{2}\).9245.

(viii) 0.000287

Characteristic: In 0.000287, there are 3 zero between decimal point and first significant digit. 

So, characteristic of its logarithm will be
-(3 + 1) = -4 = \(\bar{4}\)

Mantissa : In log table, in first column, in front of 28 (ignoring decimal point) and under the column 7 find the number which is 4579.

So, mantissa of log₁₀ 0.000287 = 0.4579

Thus,

log₁₀ 0.000287 = Characteristic + Mantissa

= \(\bar{4}\) + 0.4579

= \(\bar{4}\).4579

(ix) 1.234

Characteristic = 1- 1 = 0

Mantissa = 0.0899 + 14

= 0.0913

log₁₀1.234 = Characteristic + Mantissa

= 0 + 0.0913

= 0.0913

(x) 0.00003258

Characteristic: In 0.00003258, there are 4 zero between decimal point and first significant digit. 

So characteristic of its logarithm will be
-(4 + 1) = -5 = \(\bar{5}\)

Mantissa : In log table, in first column, in front of 32 (ignoring decimal point) and under the column 5 find the number, which is = 5119

For fourth digit 8 mean difference = 11

On adding = 5130

So, mantissa of log₁₀ 0.00003258 = 0-5130
Thus,

log₁₀ 0.00003258 = Characteristic + Mantissa

= \(\bar{5}\) + 0.5130

= \(\bar{5}\).5130

(xi) 0.000125

Characteristic : In 0.000125, there are 3 zero between decimal point and first significant digit. 

So, characteristic of its logarithm will be
– (3 + 1) = – 4 = \(\bar{4}\)

Mantissa : In log table, in first column, in front of 12 (ignoring decimal point) and under the column 5 find the number which is = 0969

So, mantissa of log₁₀0.000125 = 0.0969
Thus,

log₁₀o 0.000125 = Characteristic + Mantissa

= \(\bar{4}\) + 0.0969

= \(\bar{4}\).0969

(xii) 0.00003208

Characteristic: In 0-00003208, there are 4 zero between decimal point and first significant digit. 

So, characteristic of its logarithm will be
-(4 + 1) = -5 = 5

Mantissa : In log table, in first column, in front of 32 (ignoring decimal point) and under the column 0 find the number which is = 5051

For fourth digit 8’s mean difference = 11

On adding = 5062

So, mantissa of log₁₀0.0003208 = 0.5062
Thus,

log₁₀0.0003208 = Characteristic + Mantissa = \(\bar{5}\) + 0.5062

= \(\bar{5}\) .5062

log(11³/(5⁷ x 7⁵))

= log (11)³ – log{(5)⁷ x (7)⁵}

= 3 log 11 – log (5)⁷ – log (7)⁵

= 3 log 11 – 7 log 5 – 5 log 7

(c) 32

Hint.

Given, log₄ (2 × 4 × \(x\) × 16) = 6 ⇒ log₄ (128\(x\)) = 6 

⇒ 128x\(x\) = 4⁶              [Using log_a x = n ⇒ x = aⁿ]

log x – log (x – 1) = log 3

⇒ log(x/(x - 1)) = log 3

On comparing,

x/(x - 1) = 3

⇒ x = 3 (x – 1)

⇒ x = 3x – 3

⇒ 3x – x = 3

⇒ 2x = 3

⇒ x = 3/2

Hence, the value of x is 3/2.

(b) - 3

Hint. 

Since \({4\sqrt[3]{4^2}}\) = 2².(2⁴)^{\(^{\frac{1}{3}}\)}= \(2^{2+\frac{4}{3}}\) = \(2^{\frac{10}{3}}\) and 

1024 = 2¹⁰, therefore,

log\(_{4.\sqrt[3]{4^2}}\) \(\bigg(\frac{1}{1024}\bigg)\) = log_2^10/3 (2^-10) = \(\frac{-10}{\frac{10}{3}}\) log₂ 2

= – 3 × 1 = – 3. \(\bigg[\)Using log_a x^m = \(\frac{m}{n}\) log_a x\(\bigg].\)

(b) 1

log₄(x – 1) = log₂(x – 3) ⇒ log_2² (x − 1) = log₂(x – 3)

⇒ \(\frac{1}{2}\) log₂ (x–1) = log₂ (x– 3) ⇒ log₂ (x–1) = 2 log₂ (x– 3)

\(\big[\)Using log_a^m (bⁿ) = \(\frac{n}{m}\) log_a b\(\big]\)

⇒ log₂(x – 1) = log₂(x – 3)² 

⇒ (x – 1) = (x – 3)² ⇒ x – 1 = x² – 6x + 9 

⇒ x² – 7x + 10 = 0 ⇒ (x – 2) (x – 5) = 0 ⇒ x = 2 or 5 

Neglecting x = 2 as log₂(x – 3) is defined when x > 2.

⇒ There is only one meaningful solution of the given equation.

(b) 1 – log₂5 

Given, log₂ (5.2^x + 1), log₄ (2^{1– x} + 1), 1 are in A.P. 

⇒ log₂ (5.2^x + 1) + 1 = 2 log₄ (2^{1 – x} + 1) 

⇒ log₂ (5.2^x + 1) + log₂2 = 2 log_2² (2^1-x + 1)

⇒ log₂ (5.2^x + 1).2 = 2 x \(\frac12\)log₂ (2^1-x + 1)

\(\big(\because\,\text{log}_{a^n}x=\frac{1}{n}\text{log}_ax\big)\)

⇒ log₂ (10.2^x + 2) = log₂ (2^1-x + 1)

⇒ 10.2^x + 2 = 2^{1 – x} + 1 ⇒ 10.2^x + 2 = \(\frac{2}{2^x}\) + 1

Let 2^x = a, then

10. a + 2 = \(\frac{2}{a}\) + 1 ⇒ 10a + 1 = \(\frac{2}{a}\) ⇒ 10 a² + a – 2 = 0

⇒ (5 a – 2) (2a + 1) = 0 ⇒ a = \(\frac{2}{5}\) ⇒ 2^x = \(\frac{2}{5}\)

\(\big(\because\,2^x>0,\text{reject}\,a=-\frac{1}{2}\big)\)

⇒ log 2^x = log \(\frac{2}{5}\)

⇒ x log₂ 2 = log₂ 2 – log₂ 5 ⇒ \(x\) = 1 – log₂ 5.

(d) \(\frac{3}{2}\)

  \(2^{log_{10}3\sqrt3}\) = \(3^{k\,log_{10}}\)² ⇒ \(2^{log_{10}\big(3^{\frac{3}{2}}\big)}\) = \(3^{k\,log_{10}}\)²

⇒\(2^{log_{2}\big(3^{\frac{3}{2}}\big).log_{10^2}}\) =  \(3^{k\,log_{10}}\)2       [Using log_ax = log_bx .log_ab]

⇒ \(\bigg[2^{log_{2}\big(3^{\frac{3}{2}}\big)}\bigg]^{log_{10^2}}\) =  \((3^{k})^{log_{10}}\)² ⇒ \(2^{log_2\,3^{\frac{3}{2}}}\) = 3^k

⇒\(3^{\frac{3}{2}}\) = 3^k ⇒ k = \(\frac{3}{2}\)               (∵ \(a^{log_a\,x}\) = x)

(c) 24

Using the formula log_aⁿ x^m = \(\frac{m}{n}\) log_a x , we have

 log_√b a \(\text{log}_\sqrt[3]{c}\)  b \(\text{log}_\sqrt[4]{c}\) c = \(\text{log}_{b^\frac{1}{2}}\) a \(\text{log}_{c^\frac{1}{3}}\) b \(\text{log}_{a^\frac{1}{4}}\)c

= 2 log_b^a × 3 log_c^b × 4 log_a^c

= 24 \(\frac{log\,a}{log\,b}\times\frac{log\,b}{log\,c}\times\frac{log\,c}{log\,a}\) = 24.

log₃ [log₂(log₃ x)] = 1

\(\therefore\) log₂ (log₃ x) = 3¹

\(\therefore\) log₃ x = 2³

\(\therefore\) log₃ x = 8

\(\therefore\) x = 3⁸

\(\therefore\) x = 6561

We know that log_m mⁿ = n log_mm

and log_mm = 1

∴ log_m(m)_n = n x log_mm = n x 1 = n

⇒ log_m(m)ⁿ = n …(i)

L.H.S. = log₄[log₂{log₂(log₃ 81)}]

= log₄[log₂ {log₂(log₃3⁴)}] (∵ 81 = 3⁴)

= log₄[log₂ {(log₂⁴)} [According to equal (i)]

= log₄{log₂(log₂2²)} (∵ 4 = 2²)

= log₄(log₂2) [According to equal (i)]

= log₄(1) (∵ log_mm = 1)

= 0 = R.H.S.

Given, 1000! = n. Now, \(\frac{1}{log_2\,n}\) + \(\frac{1}{log_3\,n}\) + .... + \(\frac{1}{log_{1000}\,n}\) 

= log_n 2 + log_n 3 + .... + log_n 1000 \(\bigg[\text{Using}\frac{1}{log_b\,a}= log_a\,b\bigg]\)

= log_n (2 × 3 × 4 × ... × 1000) = log_n (1000!) = log_n n = 1.