goyalsau wrote:
y ^ 2y < y ^ y ^ 2
Base is same y so
2y < y ^ 2
I did this way , But i am not sure whether my approach is correct or not.
Please guys Limestone and fabio
Share your view on my solution.
No Sir, not good!
Let us understand this with care:
when you go from (say) x> y to a^x > a^y you are using the exponential function f(z) = a^z in the following sense:
x > y you would like to imply that f(x) > f(y) that is, a^x > a^y , correct?
So, the question is: when x> y implies f(x) > f(y) ?
Answer: when the function f is strictly increasing, that is, for greater values (in the domain), you get greater values as their images...
The fact is that f(z) = a^z is strictly increasing when a is any constant GREATER THAN 1, therefore you could use this property as limestone did BECAUSE you know that the parameter, I mean, the constant
a (in the problem´s case
y) is greater than 1, because it is greater than 2 (statement (2)).
Just to be sure you got it: try 3 > 2 then (1/2)^3 > (1/2) ^2 (MISTAKE!) to see how the parameter be greater than 1 is important.
For the sake of completeness, when the parameter
a is between 0 and 1, the function is strictly decreasing, that is, when x > y then certainly a^x < a^y ... therefore the case with (1/2) above is not a single "exception", it is a good representant of all cases where the parameter is between 0 and 1.
Crystal Clear?
Regards,
Fabio.
