BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
1)for all a>1 this is true...no issues heremetallicafan wrote: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!
It doesn't hold true for -1 < a < 0.'manpreet singh wrote:
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
Yes you are right, i missed on the odd even case here...jcnasia wrote:It doesn't hold true for -1 < a < 0.'manpreet singh wrote:
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
For example, (-.5)^2 > (-.5)^3 but 2 < 3.
New here Create free account
