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July 26th, 2016, 02:43 PM   #1
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How does one do dis???

A and B please I don't get how they're supposed to be done???
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July 26th, 2016, 03:16 PM   #2
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Can you write it out or at least make sure the image is large enough to read (and right side up?)

-Dan
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July 26th, 2016, 03:32 PM   #3
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I would start a) by determining the value of each of the logarithms.
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July 26th, 2016, 04:15 PM   #4
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Log base 2 of 2 times Log Base 2 of 4 times Log Base 2 of 8 ••• log base 2 of 2^10 is a find the exact value and
B is Log base 2 of 4 times log base 4 of 6 times log base 6 of 8 ••• log base 14 of 16 find the exact value
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July 26th, 2016, 08:41 PM   #5
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$\displaystyle \prod_{i=1}^{10} log_2(2^i) = \prod_{i=1}^{10} \{ilog_2(2)\} = \prod_{i=1}^{10} (i * 1) = \prod_{i=1}^{10} i = what?$

The pi means to multiply as many times as indicated.

There is nothing hard about this problem, but it is cumbersome without knowing a convenient notation.
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July 27th, 2016, 03:17 AM   #6
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Now what about the productmation (I'm not even sure that's what you call it, of B when the log base changes?
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July 27th, 2016, 04:16 AM   #7
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$\log_2(4) \cdot \log_4(6) \cdot \log_6(8 ) \cdot ... \cdot \log_{14}(16)$

Change of base ... let all the logs in the next expression be base 2

$\dfrac{\log(4)}{\log(2)} \cdot \dfrac{\log(6)}{\log(4)} \cdot \dfrac{\log(8 )}{\log(6)} \cdot \, ... \, \cdot \dfrac{\log(16)}{\log(14)}$

this will simplify to $4$ ... do you see why?
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July 27th, 2016, 07:11 AM   #8
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This is skeeter's answer in the pi notation.

Note that $log_2(4) = \dfrac{log_2(4)}{1} = \dfrac{log_2(4)}{log_2(2)}.$

$\displaystyle \prod_{i=1}^7 log_{2i}\{2(i+1)\} = \prod_{i=1}^7 \dfrac{log_2\{2(i+1)\}}{log_2(2i)}= \dfrac{\displaystyle \prod_{i=1}^7 log_2 \{2(i + 1)\}}{\displaystyle \prod_{i=1}^7 log_{2} (2i)} =$

$\dfrac{\displaystyle \prod_{i=2}^8log_2(2i)}{\displaystyle \prod_{i=1}^7 log_2(2i)} = \dfrac{log_2(8 * 2) * \cancel{\displaystyle \prod_{i=2}^7 log_2(2i)}}{log_2(2 * 1) * \cancel{\displaystyle \prod_{i=2}^7 log_2(2i)}} = \dfrac{log_2(16)}{log_2(2)} = 4.$

In this case, skeeter's notation is more intuitive.
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July 27th, 2016, 06:03 PM   #9
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Quote:
Originally Posted by JeffM1 View Post
This is skeeter's answer in the pi notation.

Note that $log_2(4) = \dfrac{log_2(4)}{1} = \dfrac{log_2(4)}{log_2(2)}.$

$\displaystyle \prod_{i=1}^7 log_{2i}\{2(i+1)\} = \prod_{i=1}^7 \dfrac{log_2\{2(i+1)\}}{log_2(2i)}= \dfrac{\displaystyle \prod_{i=1}^7 log_2 \{2(i + 1)\}}{\displaystyle \prod_{i=1}^7 log_{2} (2i)} =$

$\dfrac{\displaystyle \prod_{i=2}^8log_2(2i)}{\displaystyle \prod_{i=1}^7 log_2(2i)} = \dfrac{log_2(8 * 2) * \cancel{\displaystyle \prod_{i=2}^7 log_2(2i)}}{log_2(2 * 1) * \cancel{\displaystyle \prod_{i=2}^7 log_2(2i)}} = \dfrac{log_2(16)}{log_2(2)} = 4.$

In this case, skeeter's notation is more intuitive.
I'm just an "intuitive" kind of guy ...
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