Elitmus
Exam
Numerical Ability
Log and Antilog
Q)f(x)=log(base a)x + log(base b)x;where a is greater than 1 and less than b and value of log(base a)x>1,then what is the minimum value of f(x)
optn:-0,1,2,1+log(not remember proper option)
Read Solution (Total 8)
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- the last option was 1+log2(base3)..ans this ll be the ans..bcoz it can't be 1 and 1+log2(base3)
- 9 years agoHelpfull: Yes(8) No(0)
- my ans came as 2 .
- 9 years agoHelpfull: Yes(1) No(7)
- 1
log(base a)x >1
x > 1^a
min value f(x) = f(0) = 1^0=1 - 9 years agoHelpfull: Yes(1) No(3)
- f(x)=logaX+logbX
=logaX+logaX/logab(base a and argmnt b)
=logax*(1+1/logab)
=logax*((logab+logaa)/logab)
=logax*(loga(b(a-1))
acc to qusn logax >1
so (1+)*(always greter then 1(loga(b(a-1)))
so minimum value will be 1
- 9 years agoHelpfull: Yes(1) No(2)
- given 1
- 9 years agoHelpfull: Yes(0) No(0)
- 2 is the ans
- 9 years agoHelpfull: Yes(0) No(0)
- 0
differentiate
- 9 years agoHelpfull: Yes(0) No(1)
- Ans is 2.
- 8 years agoHelpfull: Yes(0) No(0)
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