100 x log x base 3 - log sqrt x base 9 X is Always greater than 1 Option a

Let 3^a = x, then 9^(a/2) = x.
Taking square root from both sides we get, (9^a/2)^1/2 = sqrt(x),
=> 9^(a/4) = srt(x).
So, log of sqrt(x) to the base 9 will be a/4.

Now the given expression will evaluate to 100^x(a - a/4) = 100^x * 3a/4.
As x is greater than 1, and we have 3^a = x, and as 3^0 = 1, a will always be greater than 0.
Also, as a ->0, expr. ->0 and as a->infinity, expr. ->infinity.
So, the required range will be (a) 0, infinity.
10 years agoHelpfull: Yes(4) No(2)
100^x(log 9 base x - log root3 base x)

= 100^x(log 3^2 base x - log 3^(1/2) 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.as x inc, value of given exp tends to infinty.

value of 100^x (log3 basex - log root9 basex) lies in interval (0, infinity)
10 years agoHelpfull: Yes(1) No(3)
@rakesh where I am doing a mistake?? help!!!!
100^x(log x(base 3)-log sqrt x(base 3^2)
= 100^x(log x(base 3)-1/4 log x base3
=100^x((log x(base 3)-1/4 log x(base 3))
=100^x((3/4 log x(base 3))
=now as x=>-3 we get -ve values...while x>=3 +ve
10 years agoHelpfull: Yes(0) No(2)

100^x (log x (base 3) - log sqrt x (base 9))... X is Always greater than 1
Option
a) 0, infinity
b) -infinity, infinity
c) infinity, -infinity
d) None