hello how would i show that the equation $\displaystyle lnx = 4x$ has one real root..
am not sure what to do with the $\displaystyle lnx$ function
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hello how would i show that the equation $\displaystyle lnx = 4x$ has one real root..
am not sure what to do with the $\displaystyle lnx$ function
You want the zeros of $\displaystyle f(x) = \ln x + x 4$, $\displaystyle 0<x<+\infty$. Observe by differentiation that f is strictly increasing. Therefore if f has at most one root, since if a<b were two distinct roots we could chuse a<c<b, and 0=f(a)<f(c)<f(b)=0, a contradiction. To see that one root exists, evaluate f near zero to get a negative value and then above 4 to get a positive value and use the IVT.
Hello, sigma1!
How about a graphing approach?
Quote:
How would i show that the equation $\displaystyle \ln x \:=\: 4x$ has one real root?
Graph the functions: .$\displaystyle \begin{Bmatrix}y &=& \ln x \\ y &=& 4x \end{Bmatrix} $ . and note their intersection(s), if any.Code:
* 
4o
 * *
 * *
 * *
 ◊
 * *
  + o    o          
 *1 4 *
 *
*
