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$f(x) = \log x$ for any real number $x > 0$ and

$$g(n)=\begin{cases} n& \text{if $n$ is even}\\ \tfrac{1}{n}& \text{if $n$ is odd}.\end{cases}$$

for any natural number $n$.

If $x$ is a natural number greater than $1$, then what is the value of

$$f(f(x^{g(10)})) – f(f(x^{g(9)})) + f(f(x^{g(8)})) –\cdots – f(f(x^{g(1)}))$$

What is the best way to this?

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  • 1
    $\begingroup$ This is not about the value of an equation at all. It's about the value of an expression. I don't know why using these words in standard ways is such a challenge to so many people. $\endgroup$ – Michael Hardy Nov 27 '11 at 16:13
  • $\begingroup$ my mistake.thanks for correcting. $\endgroup$ – sunny Nov 27 '11 at 16:32
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y(n) = f(f(x^g(n)))=f(g(n) * log(x))=log(g(n))+log(log(x))

y(n) = log(n) + log(log(x)) if n is even
y(n) = log(1/n) + log(log(x)) = -log(n) + log(log(x)) if n is odd

now,
y(10) - y(9) = log(10) + log(log(x)) + log(9) - log(log(x)) = log(10) +log (9)
similarly,
y(8) - y(7) = log(8) + log(7)

finaly result y = log(10) + log(9) + ... + log(1) = log(10*9*...*1) = log(10!)
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$$f(f(x^{g(n)}))=f(\log x^{g(n)})=f(g(n)\cdot \log x)=\log(g(n))+\log(\log x)$$

So ,

$f(\log x^{g(10)})=\log 10 +\log(\log x)$

$f(\log x^{g(9)})=\log {\frac{1}{9}} +\log(\log x)$ ...etc.

Therefore your expression is equivalent to the :

$\log 10 -\log {\frac{1}{9}}+\cdots+\log 2-\log 1$

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  • $\begingroup$ @Swapan,I can't see mistake...which line is incorrect ? $\endgroup$ – Peđa Terzić Nov 27 '11 at 9:20
  • $\begingroup$ Extremely sorry...I read the question with all + signs :( My bad, your answer is completely fine. So, I removed my wrong comment. $\endgroup$ – Tapu Nov 27 '11 at 13:01
  • $\begingroup$ @Swapan,It's ok... $\endgroup$ – Peđa Terzić Nov 27 '11 at 13:15

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