# Exponential equation

#### Albert

##### Well-known member
$f(x)=\dfrac {4^x}{4^{x+2}}$

find :$f(\dfrac{1}{2007})+f(\dfrac{2}{2007})+f(\dfrac{3}{2007})+----+ f(\dfrac{2006}{2007})$

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#### topsquark

##### Well-known member
MHB Math Helper
Re: exponential equation

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

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

#### MarkFL

Staff member
Re: exponential equation

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

$$\displaystyle \frac{1}{2007}\sum_{k=1}^{2006}k=\frac{2006\cdot2007}{2\cdot2007}=1003$$

#### Albert

##### Well-known member
Re: exponential equation

sorry a typo happened $f(x)=\dfrac {4^x}{4^{x+2}}$
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

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#### MarkFL

Staff member
Re: exponential equation

...

f(x) is not equal to $\dfrac {1}{16}$
I beg to differ:

$$\displaystyle f(x)=\frac{4^x}{4^{x+2}}=\frac{4^x}{4^x\cdot4^2}= \frac{1}{4^2}=\frac{1}{16}$$

#### Albert

##### Well-known member
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 :$f(\dfrac{1}{2007})+f(\dfrac{2} {2007})+f(\dfrac{3}{2007})+----+ f(\dfrac{2006}{2007})$

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#### mathworker

##### Active member
Re: exponential equation

for x<1003,
$$\displaystyle f(x)=\frac{1}{1+2^{1-2x}}$$
$$\displaystyle \sum f(x)=\frac{1}{1+2^{1/2007}}+\frac{1}{1+2^{3/20007}}+...........+\frac{1}{1+2^{2005/2007}}$$
for x>1003,
$$\displaystyle f(x)=\frac{2^{2x-1}}{1+2^{2x-1}}$$
$$\displaystyle \sum f(x)=\frac{2^{1/2007}}{1+2^{1/2007}}+\frac{2^{3/20007}}{1+2^{3/20007}}+...........+\frac{2^{2005/2007}}{1+2^{2005/2007}}$$
hence,

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#### Albert

##### Well-known member
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 .

#### Albert

##### Well-known member
Re: exponential equation

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

find :$f(\dfrac{1}{2007})+f(\dfrac{2} {2007})+f(\dfrac{3}{2007})+----+ f(\dfrac{2006}{2007})$
note we have :
$f(x)+f(1-x)=\dfrac {4^x}{4^x+2}+\dfrac {4^{(1-x)}}{4^{(1-x)}+2}=1$
$\therefore \,\, f(\dfrac{1}{2007})+f(\dfrac{2006}{2007})=1$
$\,\,\,\,\,\,\,f(\dfrac{2}{2007})+f(\dfrac{2005}{2007})=1$
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$\,\,\,\,\,\,f(\dfrac{1003}{2007})+f(\dfrac{1004}{2007})=1$
all together we have 1003 pairs ,so the answer is 1003