Solve Exponential Eqn: $f(x)=\dfrac {4^x}{4^{x+2}}$

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Discussion Overview

The discussion revolves around solving the exponential equation \( f(x) = \frac{4^x}{4^{x+2}} \) and its implications for a series of evaluations at fractional inputs. Participants explore the function's behavior, its simplifications, and the summation of its values over a specified range.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the function \( f(x) = \frac{4^x}{4^{x+2}} \) and seeks to evaluate the sum \( f\left(\frac{1}{2007}\right) + f\left(\frac{2}{2007}\right) + \ldots + f\left(\frac{2006}{2007}\right) \).
  • Another participant questions the purpose of \( f(x) \) in the context of the summation, suggesting a need for clarification.
  • A participant claims that \( f(x) = \frac{1}{16} \) based on simplification, which is later challenged by another participant who argues against this conclusion.
  • Subsequently, a different function \( f(x) = \frac{4^x}{4^x + 2} \) is introduced, prompting further evaluation of the same summation.
  • One participant notes that both functions yield the same results despite being different, indicating an interesting property of the functions.
  • Another participant derives a relationship \( f(x) + f(1-x) = 1 \) and uses it to conclude that the sum of pairs of function evaluations equals 1003, although this conclusion is not universally accepted.

Areas of Agreement / Disagreement

Participants express differing views on the value of \( f(x) \) and its implications. There is no consensus on the correctness of the simplifications or the conclusions drawn from the evaluations.

Contextual Notes

Some participants' claims depend on specific interpretations of the function, and there are unresolved mathematical steps regarding the summation process and the properties of the functions involved.

Albert1
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$f(x)=\dfrac {4^x}{4^{x+2}}$

find :$f(\dfrac{1}{2007})+f(\dfrac{2}{2007})+f(\dfrac{3}{2007})+----+
f(\dfrac{2006}{2007})$
 
Last edited:
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Re: exponential equation

Albert said:
$f(x)=\dfrac {4^x}{4^{x+2}}$

find :$\dfrac{1}{2007}+\dfrac{2}{2007}+\dfrac{3}{2007}+----+
\dfrac{2006}{2007}$
Okay, I get the whole add the fraction thing, but what is the f(x) for?

-Dan
 
Re: exponential equation

$$f(x)=\frac{1}{16}$$

$$\frac{1}{2007}\sum_{k=1}^{2006}k=\frac{2006\cdot2007}{2\cdot2007}=1003$$
 
Re: exponential equation

sorry a typo happened :o
$f(x)=\dfrac {4^x}{4^{x+2}}$
find :$f(\dfrac{1}{2007})+f(\dfrac{2}{2007})+f(\dfrac{3}{2007})+----+
f(\dfrac{2006}{2007})$
the original post has been edited as mentioned above
 
Last edited:
Re: exponential equation

Albert said:
...

f(x) is not equal to $\dfrac {1}{16}$

I beg to differ:

$$f(x)=\frac{4^x}{4^{x+2}}=\frac{4^x}{4^x\cdot4^2}= \frac{1}{4^2}=\frac{1}{16}$$
 
Re: exponential equation

yes, you are right ,and your solution is very smart !

now this time f(x) is different

$f(x)=\dfrac {4^x}{4^x+2}$

find :$f(\dfrac{1}{2007})+f(\dfrac{2}

{2007})+f(\dfrac{3}{2007})+----+

f(\dfrac{2006}{2007})$
 
Last edited:
Re: exponential equation

for x<1003,
$$f(x)=\frac{1}{1+2^{1-2x}}$$
$$\sum f(x)=\frac{1}{1+2^{1/2007}}+\frac{1}{1+2^{3/20007}}+...+\frac{1}{1+2^{2005/2007}}$$
for x>1003,
$$f(x)=\frac{2^{2x-1}}{1+2^{2x-1}}$$
$$\sum f(x)=\frac{2^{1/2007}}{1+2^{1/2007}}+\frac{2^{3/20007}}{1+2^{3/20007}}+...+\frac{2^{2005/2007}}{1+2^{2005/2007}}$$
hence,
answer is 1003
 
Last edited:
Re: exponential equation

It is very interesting ,we have two different functions ,
but their answers are the same ,this is what I want to
point out .
my solutions are different ,and I will post it later .
 
Re: exponential equation

Albert said:
$f(x)=\dfrac {4^x}{4^x+2}$

find :$f(\dfrac{1}{2007})+f(\dfrac{2}

{2007})+f(\dfrac{3}{2007})+----+

f(\dfrac{2006}{2007})$
note we have :
$f(x)+f(1-x)=\dfrac {4^x}{4^x+2}+\dfrac {4^{(1-x)}}{4^{(1-x)}+2}=1$
$\therefore \,\, f(\dfrac{1}{2007})+f(\dfrac{2006}{2007})=1$
$\,\,\,\,\,\,\,f(\dfrac{2}{2007})+f(\dfrac{2005}{2007})=1$
-------------------------------------------------------------------------
-------------------------------------------------------------------------
$\,\,\,\,\,\,f(\dfrac{1003}{2007})+f(\dfrac{1004}{2007})=1$
all together we have 1003 pairs ,so the answer is 1003
 

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