Find the value of '$x$' if, $$\large \left(\frac{1}{2^{\log_x 4}} \right) \cdot \left( \frac{1}{2^{\log_x 16}} \right) \cdot \left(\frac{1}{2^{\log_x 256}} \right) \cdots = 2 $$

I tried to make it simple by resulting series is not converging, the suggested answers in my module is either ($2,\frac{1}{2},4,\frac{1}{4}$)

Also I tried this and this but not the correct answer in either case.

  • $\begingroup$ Did you perhaps mean $\log_x{4}, \log_x{16}, etc$ as the exponents? $\endgroup$ – Tyler Mar 19 '11 at 5:36
  • $\begingroup$ What does $\log_x^y$ mean? Presumably the sequence $4,16,256$ is continued by squaring. In other words, the sequence is $\displaystyle{4^{2^k}}$ for $k=0,1,2,\ldots$? $\endgroup$ – Jonas Meyer Mar 19 '11 at 5:37
  • $\begingroup$ @Debanjan - what you have written does not make sense. If you mean $\log_x 4,\log_x 16, \cdots$ then note that $4=2^2$, $16 = 2^4$, $256 = 2^8$ $\endgroup$ – Juan S Mar 19 '11 at 5:49
  • $\begingroup$ @Jonas Meyer: $\log_x^y$ means logarithm of $y$ base $x$. $\endgroup$ – Quixotic Mar 19 '11 at 5:50
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    $\begingroup$ The terms in the product converge to $0$ if $x>1$ and to $\infty$ if $0<x<1$, so the equality is impossible for all $x$. $\endgroup$ – Jonas Meyer Mar 19 '11 at 5:59

I'm going to suggest that the "correct" answer is 4 and that the reason why is that there are two mistakes in the problem. x=4 is the correct answer to the question: Find the value of '$x$' if, $$\large \left(\frac{1}{2^{\log_x 4}} \right) + \left( \frac{1}{2^{\log_x 16}} \right) + \left(\frac{1}{2^{\log_x 64}} \right) + \cdots = 2 $$

So, I'm suggesting that it should both have been a sum rather than a product (since in its current form, it has no solution as Jonas correctly pointed out) and that the increasing series should have been $4^i$ rather than $4^{2^i}$. Although it can be solved if we just change it to a sum and leave it as $4^{2^i}$, it's difficult to solve and won't come out evenly (using Wolfram to calculate, the correct answer for just changing it to a sum is greater than 63, but less than 64). But if we change it to be 4,16,64,256,..., then we get the nice tidy answer of 4. This is probably what they were looking for, but they completely botched the question.


$$ \begin{align} \left(\frac{1}{2^{\log_x 4}}\right)\left(\frac{1}{2^{\log_x 16}}\right)\left(\frac{1}{2^{\log_x16}}\right)\cdots&=2\\ 2^{-\log_x 4}\times2^{-\log_x 16}\times2^{-\log_x 256}\ldots&=2\\ 2^{-\left(\log_x 4+\log_x 16+\log_x 256+\cdots\right)}&=2\\ \log_x(4\times16\times256\times\cdots)&=-1\\ 4\times 16\times256\times\cdots&=\frac{1}{x}\\ 2^2\times2^4\times2^8\times\cdots&=\frac{1}{x}\\ 2^{2+4+8+\cdots}&=\frac{1}{x} \end{align} $$ Now say we have, $2+4+8+\cdots$; this can be written as $2+4+8+\cdots+2^n$ when $n\to\infty$. $2+4+8+\cdots+2^n$ is a geometric series and: $$2+4+8+\cdots+2^n=2\left(2^n-1\right)$$ So now we have: $$ \begin{align} x&=\lim_{n\to\infty}\left(\frac{1}{2^{2\left(2^n-1\right)}}\right) \end{align}$$


Hint: $\log_x 4 = log_x 2^2 = 2 log_x 2$

$\log_x 16 = log_x 2^4 = 4 log_x 2$

Let $y = log_x 2$ and see what you get

Edit - Jonas is correct

The product is

$$\lim_{n \to \infty} \prod_{k=1}^n 2^{n(n+1)(-y)}$$

  • $\begingroup$ so the series is going to be $- 2 log_x 2 - 4 log_x 2 - 8 log_x 2 - \cdots = 1 $ but again this is not converging to me ... or am I doing some mistake? $\endgroup$ – Quixotic Mar 19 '11 at 6:05
  • $\begingroup$ I really don't understand. $\endgroup$ – Quixotic Mar 19 '11 at 6:20
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    $\begingroup$ @Debanjan: There is no solution, as I commented on your question earlier. If $x>1$ then $\log_x(z)\to\infty$ as $z\to\infty$, so the terms in the product go to $0$, and therefore the product is $0$. If $0<x<1$, then $\log_x(z)\to-\infty$ as $z\to\infty$, so the terms in the product go to $\infty$, and therefore the product diverges to $+\infty$. $\endgroup$ – Jonas Meyer Mar 19 '11 at 6:25

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