If $\sum_{r=0}^{n-1}\log _2\left(\frac{r+2}{r+1}\right)= \prod_{r = 10}^{99}\log _r(r+1)$, then find $n$.

Question

If \begin{align}\sum_{r=0}^{n-1}\log _2\left(\frac{r+2}{r+1}\right) = \prod_{r = 10}^{99}\log _r(r+1).\end{align} then find $n$.

I found this question in my 12th grade textbook and I just can't wrap my head around it. I tried splitting the logarithms by the division rule, which didn't work so I'm out of ideas.

Answer 1

It is: $$\sum_{r=0}^{n-1}\log _2\left(\frac{r+2}{r+1}\right) = \log_2 \frac21+\log_2\frac23+\cdots+\log_2\frac{n+1}{n}=\\ \log_2\frac21\cdot\frac32\cdots\frac{n+1}{n}=\log_2(n+1)\\ \prod_{r = 10}^{99}\log _r(r+1)=\log_{10}11\cdot\log_{11}{12}\cdots\log_{99}{100}=\\ \frac{\log_211}{\log_210}\cdot\frac{\log_212}{\log_211}\cdots\frac{\log_2100}{\log_299}\cdot=\frac{\log_2100}{\log_210}=2$$ Can you find $n$? Wolfram answer.

Answer 2

Hint:

If $f(n)=\log_an$ is defined for $n=r+2,r+1$

$$\log_a\dfrac{r+2}{r+1}=f(r+2)-f(r+1)$$

Using Telescoping Series, $$\sum_{r=0}^{n-1}\log_a\dfrac{r+2}{r+1}=f(n-1+2)-f(0+1)=?$$

Again for the right hand side, $$\log_r(r+1)=\dfrac{\log_br+1}{\log_br}=\dfrac{g(r+1)}{g(r)}$$ where $g(m)=\log_bm$ is defined for $m=r+1,r$

$$\prod_{r=10}^{99}\log_r(r+1)=\dfrac{g(99+1)}{g(10)}=?$$

Answer 3

I agree with the other answers. However, if I was fairly new to logarithms, I would have approached the problem differently.

$\log_2 \,\frac{a}{b} \;=\; log_2 \,a \;-\; log_2 \,b.$
Therefore, the LHS is easily seen to be a telescoping series.

At this point, I would regard the RHS as seemingly ugly, and I would metacheat. That is, I would guess that a moderate exploration of the RHS will reveal a pattern.

Therefore, I would begin exploring:
Let $a_n \;\equiv\; \log_n \,(n+1).$
Then the RHS = $a_{10} \times a_{11} \times a_{12} \times \cdots \times a_{99}.$

Looking for a pattern, I would see that
$10^{(a_{10})} = 11\;$ and $\;11^{(a_{11})} = 12.$
Then, I would try to evaluate $10^{(a_{10}) \times (a_{11})}.$

The key to deciphering the RHS is to realize that the above equals
$\left[10^{(a_{10})}\right]^{(a_{11})} \;=\; 11^{(a_{11})} = 12.$
Using similar analysis, I would find that $10^{(a_{10}) \times (a_{11}) \times (a_{12})} = 13.$

Let $P \;\equiv\; a_{10} \times a_{11} \times a_{12} \times \cdots \times a_{99}\;$ (i.e. $P$ = RHS).
Then, in order to solve the problem, you must use the previous analysis to conclude that $10^P = 100,\;$ so $P = 2.$

Given that the answers provided by other posters are accurate, the point of my posting is:

(1) Faced with an ugly looking problem, metacheat. That is, assume that the problem will yield to fairly straightforward analysis that involves the concepts (e.g. concepts around the logarithm) that lead up to this problem. Otherwise, why would you be confronted with this problem at this time?

(2) Faced with an ugly looking problem, don't swing for the fences. Instead, take baby steps, look for a pattern in the analysis, and see if such a pattern can be applied to the overall problem. For the present problem, the baby step is to try to evaluate $\;10^{(a_{10}) \times (a_{11})}.$