A tricky problem from Oxford Entrance Exams || No Calculator Allowed 📵

[@borismarinkovicgutierrez2349]

Can be solved only with a logarithmic table, basic arithmetics and basic mathematical logic. 16 log 18 < 18 log 16. Since logarithmic function is always progresive, then 18^16 < 16^18.

[@toshimakusugamo]

18^6 v.s 16^18
16^18 / 18^16
= 16^16 / 18^16 * 16^2
= ( 16 / 18 )^16 * 16^2
= ( 8 / 9 )^16 * 2^8
= ( 2^0.5 * 8 / 9 )^16
2^0.5 > 1.96^0.5 = ( 1.4^2 )^0.5 = 1.4
→ 2^0.5 > 1.4
16^18 / 18^16
= ( 2^0.5 * 8 / 9 )^16
> ( 1.4 *8 /9 )^16
>( 9 / 9 )^16 = 1
16^18 / 18^16 > 1
16^18 > 18^16

[@rajkunargupta5183]

18 to power 16 is greater

[@CharlesChen-el4ot]

=（18/16）^16/256
16*log(18/16)/2
=8*log(9/8)>1

[@RealQinnMalloryu4]

2^94^4vs4^4^2^9 1^3^2^2^2^2^2vs2^2^2^2^1^3^2 1^1^1^1^1^2vs1^1^1^2^3^1 12vs 23 (x ➖ 2x+1). (x ➖ 3x+2)
18^16