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Let $f(x)=\ln{x}-\dfrac{1}{x}+3$, and $a_{i}>0,i=1,2,3$, such that $a_{3}:a_{2}:a_{1}=e^2:e:1$.

Suppose $f(a_{1})+f(a_{2})+f(a_{3})=\dfrac{e^5-e^2}{1-e}$; what is the value of $\dfrac{f(a_{1})+f(a_{3})}{f(a_{2})}$? (where $e=2.718\cdots, \ln{x}=\log_{e}{x}$)

A: $\dfrac{e^2+1}{e}\quad $ B:$\dfrac{e^2+3}{e+1}\quad$ C :$\dfrac{e^2+5}{e+2}\quad$ D :$\dfrac{e^3+e+2}{e^2+1}$

this problem have nice solution? Thank you

my idea:$a_{3}=e^2*a_{1},a_{2}=e*a_{1}$

since $f(a_{1})+f(a_{2})+f(a_{3})=3\ln{a_{1}}+3-\dfrac{1}{a_{1}}\left(1+\dfrac{1}{e}+\dfrac{1}{e^2}\right)+9=\dfrac{e^5-e^2}{1-e}$,

$$\Longrightarrow 3\ln{a_{1}}=\dfrac{1}{a_{1}}\left(1+\dfrac{1}{e}+\dfrac{1}{e^2}\right)+\dfrac{e^5-e^2}{1-e}-12$$ then find the value


then we must find the $a_{1}$,Now I think follow is very ugly.someone have nice methods? Thank you

oh, I have solution:let $a_{2}=x$, It's easy have $$3\ln{x}+9=(e^2+e+1)\left(\dfrac{1}{ex}-e^2\right)$$ so let $$F(x)=3\ln{x}+9-(e^2+e+1)\left(\dfrac{1}{ex}-e^2\right),x>0$$ so $F'(x)=\dfrac{3}{x}+\dfrac{e^2+e+1}{e}\dfrac{1}{x^2}>0$ and we have $F(e^{-3})=0$

so $a_{2}=e^{-3}$

and follwing is very easy

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what does the notation $a_3:a_2:a_1=e^2:e:1$ mean? – Heberto del Rio Apr 13 '13 at 1:59
mean $a_{3}=e^2*a_{1},a_{2}=e*a_{1}$ – math110 Apr 13 '13 at 2:01
So they are three successive terms of a geometric series with ratio $e$ – muzzlator Apr 13 '13 at 2:37
@muzzlator,Yes. – math110 Apr 13 '13 at 2:43
up vote 1 down vote accepted

Hint$$f=\ln e^3x -\dfrac1x$$ $$f(a_1)+f(a_2)+f(a_3)=3ln(ae^2)-\dfrac1a\bigg[1+\dfrac1e+\dfrac1{e^2}\bigg]$$


where $a$ is 1st term of the $GP:(a1,a2,a3)$

Now , $$f(a_1)+f(a_2)+f(a_3)=-\dfrac{e^5-e^2}{e-1}$$

Comparing both the equations.$$ln(ae^2)=0\ ;\ \dfrac1a\bigg[\dfrac{e^3-1}{e-1}\bigg]=\dfrac{e^5-e^2}{e-1}$$

So, $$a=\dfrac1{e^2}$$ Now we can proceed use this value for a

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