Evaluate $\log_{2005}(1/2)\log_{2004}(1/3)\log_{2003}(1/4)\ldots\log_2(1/2005)$

Question

The numbers 2005, 2004, 2003, ..., 2 are the bases. I cannot understand how to start the question. Please help. What to do in these type of questions? Thanks in advance.

Answer 1

Isn't this equal to $(-1)^{2004}$ ?

$$ \prod_{n=1}^N \log_{n+1}(\frac{1}{N+1-(n-1)}) = \prod_{n=1}^N -\log_{n+1}(N+1-(n-1)) = \prod_{n=1}^N -\frac{\log(N+1-(n-1))}{\log(n+1)}=\frac{\prod_{n=1}^N -\log(N+1-(n-1))}{\prod_{n=1}^N \log(n+1)}=(-1)^N$$

Answer 2

The way to think about questions like this is to go back to the definitions. It is very easy to get confused.

But if we let $y_1=\log_{2005}\frac 12$ then we have $2005^{y_1}=\frac 12$

Now we want $y_1$ on its own, so that we can multiply it by other expressions of the same type. So we take logs to our favourite base and obtain $$y_1\log 2005=-\log 2$$

And continuing in the same vein we see $$y_2\log 2004=-\log 3$$$$y_3\log 2003=-\log 4$$$$y_4\log 2002=-\log 5$$ending with $$y_{2004}\log 2=-\log 2005$$

We can then see quite easily that the logs will cancel, leaving only the sign to worry about - and using the index for $y_r$ tells us that there are $2004$ products and the answer is $1$.