[tex]17^{14}> 31^{11}[/tex] if and only if
[tex]14log(17)> 11log(31)[/tex] which is true if and only if [tex]\frac{14}{11}> \frac{log(31)}{log(17)}[/tex].

[tex]\frac{14}{11}= 1.272727272727272...[/tex]
and
[tex]\frac{log(31)}{log(17}= 1.21204681309626[/tex].

Hey,
has helped a lot - thanks!:)

hi

17[power]14[/power]==17[power]11[/power]*17[power]3[/power]

if you show:
17[power]11[/power]*17[power]3[/power] { } 31[power]11[/power]
and it is equivalent whit 17[power]3[/power] { } 31[power]11[/power] / 17[power]11[/power]
and this 31[power]11[/power] / 17[power]11[/power]=(31/17)[power]11[/power] [tex]\approx[/tex] 2[power]11[/power]

now you can compare 17[power]3[/power] { } 2[power]11[/power]
and this you can easy comare

17[power]3[/power]{ [b]>[/b] }2[power]11[/power]

Hi!

I know that the power of 17^14 is greater, but I do wonder how you could show it without using obviously any calculation help, but with the mere use of the laws of potency?
I just cannot find even a start and would appreciate any input and any hint.

regards,
Placebo