Home
Class 12
MATHS
Let x=(0. 15)^(20)dot Find the character...

Let `x=(0. 15)^(20)dot` Find the characteristic and mantissa of the logarithm of `x` to the base 10. Assume `log_(10)2=0. 301 and log_(10)3=0. 477.`

Text Solution

Verified by Experts

` log x = log(0.15)^(20)= 20 log (15/100)`
` = 20 [log 15-2]`
` = 20[ log 3+log 5-2]`
` = 20[log 3+1-log 2 - 2]" "(.:' log_(10) 5 = log_(10) 10/2)`
`=20[-1+log 3 - log 2]`
` =20 [ -1+0.477-0.301]`
` =- 20 xx 0.824`
` =- 16.48`
` = bar(17).52`
Hence, Characteristic =- 17 and Mantissa = ` 0.52`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • LOGARITHM AND ITS PROPERTIES

    CENGAGE ENGLISH|Exercise ILLUSTRATION 1.80|1 Videos
  • LOGARITHM AND ITS PROPERTIES

    CENGAGE ENGLISH|Exercise ILLUSTRATION 1.81|1 Videos
  • LOGARITHM AND ITS PROPERTIES

    CENGAGE ENGLISH|Exercise ILLUSTRATION 1.78|1 Videos
  • LOGARITHM AND ITS APPLICATIONS

    CENGAGE ENGLISH|Exercise Subjective Type|9 Videos
  • MATHMETICAL REASONING

    CENGAGE ENGLISH|Exercise Archives|10 Videos

Similar Questions

Explore conceptually related problems

If mantissa of lagarithm of 719.3 to the base 10 is 0.8569 then mantissa of logarithm of 71.93 is

Express log_(10)2 + 1 in the form of log_(10)x .

[log_10⁡(x)]^2 − log_10⁡(x^3) + 2=0

Using differentials, find the approximate value of log_(10)10.1 when log_(10)e=0.4343 .

Find the value of : log_(10) 0.001

Let S denotes the antilog of 0.5 to the base 256 and K denotes the number of digits in 6^(10) (given log_(10)2=0.301 , log_(10)3=0.477 ) and G denotes the number of positive integers, which have the characteristic 2, when the base of logarithm is 3. The value of SKG is

Comprehension 3 If P is the non negative characteristic of (log)_(10)N , the number of significant digit in N\ i s\ P+1 . If P is the negative characteristic of (log)_(10)N , then number of zeros after decimal before a significant digit start are P-1.( Use log_(10)\ 2=0. 301.(log)_(10)3=0. 4771 ) Number of significant digit in N , where N=(5/3)^(100), is- a. 23 b. \ 22 c. \ 21 d. none

Evaluate: log_(10)(0.06)^(6)

Using differentials, find the approximate value of (log)_(10)10. 1 , it being given that (log)_(10)e=0. 4343 .

If log_(10)x+log_(10)y=2, x-y=15 then :