+91-87964 74404
FlagBACK
Question icon
Grade 11IIT JEE Entrance Exam

The number N = 6 log 2 to the base 10 + log 31 to the base 10 lies between two successive integers whose sum is equal to
(1) 5
(2) 7
(3) 9
(4) 10

Profile image of Ananya
7 Years agoGrade 11
Answers icon

2 Answers

Profile image of Saurabh Koranglekar
7 Years ago
log of ((2^6)
Profile image of Saurabh Koranglekar
7 Years ago
6 log of 2 to base 10 = log of 32 to base 10
 
log of 31 to base 10 be as it is
 
now log of a to base 10 + log of b to base 10 = log of (a*b) to base 10
 
therefore, log of (32*31) to base 10 = 2.9965
 
therefore ans = 2+3
                     option number (1) 5

Other Related Questions on iit jee entrance exam

question mark

Let S={x:x belongs to R(√3+√2)^(x^2-4) - (√3-√2)^(x^2-4)=10}. Then n(S) is equal to-

iit jee entrance exam
1 Answer Available

Last Activity: 10 Months ago

question mark
My rank is 29k, General, should I opt for any state college cse branch or some Nits in which in which I am getting ECE , chemical, mechanics etc....
 
iit jee entrance exam
1 Answer Available

Last Activity: 3 Years ago

question mark
How is Ranquer Eduzone,comparing, Akash,Fitjee,Resonance,Naryana
iit jee entrance exam
1 Answer Available

Last Activity: 4 Years ago

question mark
Can class 9th students study syllabus class  of 11th 
iit jee entrance exam
3 Answers Available

Last Activity: 4 Years ago

question mark
Sum of series 2[n/2] ! [n/2]!/ n! [2Co-2C2+3C2....+[-1]^n[n+1]2Cn] where n is an even positive integer. Kindly provide solution of this question and how to generalise this. How do we know that which expansion will give this?
iit jee entrance exam
1 Answer Available

Last Activity: 4 Years ago