text: # How to solve this IVT problem??
  
  • Oct 18th 2009, 01:33 PM
  skboss
  How to solve this IVT problem??
  How do I solve this problem?
  
  Use IVT (Intermediate Value Theorem) to prove the following statement:
  
  2^x = bx has a solution if b>2
  
  Attempt: I know that IVT says that a continuous function has values in every single points, but if I don't the value in front of x how am I supposed to solve that?
  • Oct 18th 2009, 01:46 PM
  HallsofIvy
  Quote:
  
  Originally Posted by skboss
  How do I solve this problem?
  
  Use IVT (Intermediate Value Theorem) to prove the following statement:
  
  2^x = bx has a solution if b>2
  
  Attempt: I know that IVT says that a continuous function has values in every single points, but if I don't the value in front of x how am I supposed to solve that?
  
  Let f(x)= \$\displaystyle 2^x- bx\$. Then f(1)= 2- b< 0. Whatever b is, there exist n such that \$\displaystyle 2^n> b\$. For that n, \$\displaystyle f(2^n)= 2^{2^n}- b(2^n)= 4^n- 2^nb= 2^n(2^n- b)> 0\$.
  
  Now use the IVT
  • Oct 18th 2009, 02:03 PM
  skboss
  Quote:
  
  Originally Posted by HallsofIvy
  Let f(x)= \$\displaystyle 2^x- bx\$. Then f(1)= 2- b< 0. Whatever b is, there exist n such that \$\displaystyle 2^n> b\$. For that n, \$\displaystyle f(2^n)= 2^{2^n}- b(2^n)= 4^n- 2^nb= 2^n(2^n- b)> 0\$.
  
  Now use the IVT
  
  Thank you very much, but I got lost after the 2nd step when u introduce n. If you can explain that, that would be great. Thanks again.
  • Oct 19th 2009, 04:19 AM
  HallsofIvy
  Why part do you need explained?
  
  Do you understand why I wanted to be able to say that f(1)< 0 and that f(some number) > 0?
  
  Do you understand why \$\displaystyle 2^{2^n}= 4^n= (2^n)(2^n)\$?