Forum
Forums
Questions and Answe...
Algebra
Formulas
Logarithm and seque...
 
Share:
Notifications
Clear all

[Solved] Logarithm and sequence

 
Formulas
Last Post by Staff 4 years ago
1 Posts
1 Users
0 Reactions
271 Views
RSS
0
June 15, 2021 10:50 am
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Let the function f(x)=log_{2}x-log_{x}2 (0<x<1),   and the sequence \left \{ a_{n} \right \}  satisfies f(2^{a_{n}})=2n.  Find a_{n}.

 

Add a comment
1 Answer
0
June 15, 2021 10:50 am
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Since  \small f(x)=log_{2}x-log_{x}2=log_{2}x-\frac{1}{log_{2}x}, then

\small f(2^{a_{n}})=log_{2}2^{a_{n}}-\frac{1}{log_{2}2^{a_{n}}}=a_{n}-\frac{1}{a_{n}}=2n.

It leads to \small (a_{n})^2-2na_{n}-1=0.

Hence \small a_{n}=n\pm \sqrt{n^2+1}.

Since  \small 0<x<1,  then \small 0<2^{a_{n}}<1  which means \small a_{n}<0. 

Therefore \small a_{n}=n-\sqrt{n^2+1}(n\in \mathbb{N}).

Add a comment
Forum Jump:
  Previous Topic
Next Topic  
Share: