Log third

Algebra Level 3

For some constant a > 1 a> 1 , the two functions f ( x ) = a x f(x) = a^x and g ( x ) = log a x g(x) = \log_a x intersect exactly once.

What is a ? a?

e 1 e^{1 } e 1 e^{ -1 } e e e^{e } e e e^{ -e } e 1 e e^{ \frac{1}{e} }

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Prince Loomba
May 1, 2016

W e k n o w t h a t t h e s e f u n c t i o n s a r e i n v e r s e o f e a c h o t h e r a n d i n v e r s e f u n c t i o n s a r e m i r r o r i m a g e s a b o u t y = x . S o w h e r e v e r t h e y i n t e r s e c t , t h e i r s l o p e s w i l l b e e q u a l t o 1. a x = l n x l n a . . . ( 1 ) ( b y b a s e c h a n g i n g t h e o r e m ) . a x l n a = 1.... ( 2 ) , 1 x l n a = 1... ( 3 ) . U s i n g t h e s e 3 e q u a t i o n s , f r o m 3 , x l n a = 1 , l n a = 1 x , e l n a = e 1 x , a = e 1 x . U s i n g 1 , ( a x = l n x l n a ) ; f r o m 2 , a x l n a = 1 , l n a = 1 a x . S u b s t i t u t i n g t h i s v a l u e i n t h e e q u a t i o n 1 , l n x = 1 , x = e . P u t t i n g x = e i n t h e o b t a i n e d e q u a t i o n , w e g e t a = e 1 e We\quad know\quad that\quad these\quad functions\quad are\quad inverse\quad of\quad each\quad other\\ and\quad inverse\quad functions\quad are\quad mirror\quad images\quad about\quad y=x.\\ So\quad wherever\quad they\quad intersect,\quad their\quad slopes\quad will\quad be\quad equal\quad to\quad 1.\\ a^{ x }=\frac { lnx }{ lna } ...(1)(by\quad base\quad changing\quad theorem).\\ a^{ x }lna=1....(2),\\ \frac { 1 }{ xlna } =1...(3).\\ Using\quad these\quad 3\quad equations,\\ from\quad 3,xlna=1,lna=\frac { 1 }{ x } ,e^{ lna }=e^{ \frac { 1 }{ x } },\boxed { a=e^{ \frac { 1 }{ x } } } .\\ Using\quad 1,(a^{ x }=\frac { lnx }{ lna } );\\ from\quad 2,a^{ x }lna=1,lna=\frac { 1 }{ a^{ x } } .\\ Substituting\quad this\quad value\quad in\quad the\quad equation\quad 1,\quad lnx=1,\quad x=e.\\ Putting\quad x=e\quad in\quad the\quad obtained\quad equation,we\quad get\quad a={ e }^{ \frac { 1 }{ e } }\\

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...