Exponents theory 2

[metallicafan]

Hi,

I would like to know in which cases we can say that because a^x > a^y, then x>y.

If a is a fraction is between 0 and 1, we cannot claim that, right?

Thank you!

[\'manpreet singh]

Hi,

I would like to know in which cases we can say that because a^x > a^y, then x>y.

If a is a fraction is between 0 and 1, we cannot claim that, right?

Thank you!

1)for all a>1 this is true...no issues here

2)for 0<a<1
let a=0.5 0.5^1>0.5^2 here the above relation is not true as 1>2 (false)

3)for -1<a<0 again let a=-0.5
-0.5^1<-0.5^2 and 1<2...the relation holds valid here
-0.5^2>-0.5^3 but 2>3...the relation is not valid anymore

4)for a<-1 a=-2 -2^1>-2^2 but here 1>2 (which is false)hence the above relation not valid


Hence the range of 'a' for which the above relation holds true is:
a>1 for all x,y
and -1<a<0 for all odd values of x and y(definitely)

======================
manpreet

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[jcnasia]

1)for all a>1 this is true...no issues here

2)for 0<a<1
let a=0.5 0.5^1>0.5^2 here the above relation is not true as 1>2 (false)

3)for -1<a<0 again let a=-0.5
-0.5^1<-0.5^2 hence 1<2...the relation holds valid here

4)for a<-1 a=-2 -2^1>-2^2 but here 1>2 (which is false)hence the above relation not valid


Hence the range of 'a' for which the above relation holds true is:
a>1 and -1<a<0

It doesn't hold true for -1 < a < 0.
For example, (-.5)^2 > (-.5)^3 but 2 < 3.

[\'manpreet singh]

1)for all a>1 this is true...no issues here

2)for 0<a<1
let a=0.5 0.5^1>0.5^2 here the above relation is not true as 1>2 (false)

3)for -1<a<0 again let a=-0.5
-0.5^1<-0.5^2 hence 1<2...the relation holds valid here

4)for a<-1 a=-2 -2^1>-2^2 but here 1>2 (which is false)hence the above relation not valid


Hence the range of 'a' for which the above relation holds true is:
a>1 and -1<a<0

It doesn't hold true for -1 < a < 0.
For example, (-.5)^2 > (-.5)^3 but 2 < 3.

Yes you are right, i missed on the odd even case here...

We can then generalize that it holds true for all odd values of x and y.(-1<a<0)

Made the correction in my post too.