Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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).
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !