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

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

log₁₀250-log₁₀(1.732)

=log₁₀250-log₁₀₍√3)

=log₁₀250-log₁₀(3)^1/2

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

=[2log₁₀250-log₁₀3]/2

=[log₁₀(250)²-log₁₀3]/2

=log₁₀[(250)²/3],and solve further.

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.

Regards,

Ashwin (IIT Madras).