Elitmus
Exam
Numerical Ability
Log and Antilog
log(3base)N +log(9base)N =
a. 1
b. 9
c. 111
d .333
Read Solution (Total 8)
-
- take n=729
log(3base)729+log(9base)729
6log(3base)3+3log(9base)9
6*1+3*1=9
so,only one value is possible for whole number...coz in question n is 3 digit number gvn.
- 10 years agoHelpfull: Yes(17) No(2)
- i think it should be 9 because i find two values of n so it is impossible that ans should be 1 so 9 will be the ans
- 10 years agoHelpfull: Yes(7) No(0)
- 1
only N=9^3 - 10 years agoHelpfull: Yes(4) No(1)
- The ans. will be 9.
- 10 years agoHelpfull: Yes(4) No(2)
- bhaiyon pehle yeh batao ki hum logon ko n ki value nikalni thi ya pure term ki
- 10 years agoHelpfull: Yes(4) No(1)
- ans.1
actually we have two posiblelites243,729
if we take 243 then ans is 5+2.5=7.5 it is not whole number,so we hav eonly one posibility.
ans will be whole number clearly mention in quetion. - 10 years agoHelpfull: Yes(3) No(2)
- 729= 3^6 ,9^3
6log3_3 + 3log9-9= 6+3 - 10 years agoHelpfull: Yes(3) No(2)
- ans will be 1
- 10 years agoHelpfull: Yes(2) No(1)
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