Latest Elitmus Aptitude Question SOLUTION: Solve 100^x(log 9 base x - log root3 base x) X is Greater then 1. a)0,Infinity b)-Infinity , Infinity c)Infinity,-Infinity d)None

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(#M40034958) ELITMUS Numerical Ability question Logarithm Keep an EYE

Solve
100^x(log 9 base x - log root3 base x)
X is Greater then 1.
a)0,Infinity
b)-Infinity , Infinity
c)Infinity,-Infinity
d)None

Asked In Elitmus tausif (10 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

RAKESH  (10 years ago)

100^x(log 9 base x - log root3 base x)

100^x(2*log3 base x - (1/2) *log3 base x)

100^x(3/2)*log3 base x

as x > 1
for any values of x, given exp is always +ve, it is never equal to zero or -Infinity.

d)None

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tausif  (10 years ago)

Hello Rakesh If the Statement in always +ve Then We can say that the Value is in between 0 to infinity?????

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souuuu  (10 years ago)

-infinity,0

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tausif  (10 years ago)

Not in the Option so u are wrong

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abhishek mishra  (10 years ago)

100^x(log 3 base x-log 9 base root x) so when we simplify this we'll get -3*100^x/log x base 3 and in the question log x base 3 was greater than 1(log x base 3>1)so obviously x would be greater than 3(x>3) no any value u put u'll get the negative value only so option does not match and
answer (d)

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sudheer kumar choudhary  (10 years ago)

100^x(log 9 base x - log root3 base x)

100^x(2*log3 base x - (1/2) *log3 base x)

100^x(3/2)*log3 base x

as x > 1
for any values of x, given exp is always +v
so the expression value will lie in between 0 and infinity..and the range does not include both the starting and end points here, so the option looks more appropriate.

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ASHISH MUKHERJEE  (9 years ago)

According to question is x>1 then it means
the obtained equation 100^x3/2ln (base 3) x
will be never 0 so option must be D.

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