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How to calc. value of log 250(base 10) - log 1.732(base 10)? Explain in detail.

How to calc. value of log 250(base 10) - log 1.732(base 10)? Explain in detail.


2 Answers

Menka Malguri
39 Points
9 years ago

Using property loga-logb=log(a/b),we can solve this ques.




=log10250-(1/2)log103..................(since logmn=nlogm)



=log10[(250)2/3],and solve further.

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Manish,


This could be a typical IIT JEE Objective type question.


You need to be absolutely clear about the properties of logarithms, to evaluate this, without using a log book.


(Here, i am not using any number in the base. Assume the base is 10)


So log(1.732) = log[root(3]) = (1/2)*log3  --------------- {Using the property log(a^b) = b*log(a)}


Now log(250) = log(25*10) = log25 + log10 ---------------- {Using the roperty log(a*b) = log(a) + log(b)}

So that is equal to log(25)+1 = log(5^2) + 1 = 1 + 2*log5 = 1 + 2*log(10/2) = 1 + 2*{1-log2} = 3 - 2*log2


So finally you are left with, 3 + (1/2)*log3 - 2*log2.


In case you have this question is IIT-JEE, the value of log(2) and log(3) to the base 10 would be given in terms of decimals. ------------ {For other cases you could remember the value of log(2) and log(3) as 0.30103 and 0.4771}


So you required value would be 3 + (1/2)*(0.4771) - 2*(0.30103) = 2.63649.


Hope that helps.



Ashwin (IIT Madras).

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