My Math Forum How does one do dis???

 Pre-Calculus Pre-Calculus Math Forum

July 26th, 2016, 02:43 PM   #1
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Math Focus: Calculus and Algebra
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 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timey-wimey stuff. Can you write it out or at least make sure the image is large enough to read (and right side up?) -Dan
 July 26th, 2016, 03:32 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra I would start a) by determining the value of each of the logarithms.
 July 26th, 2016, 04:15 PM #4 Newbie   Joined: Jun 2016 From: New York Posts: 14 Thanks: 0 Math Focus: Calculus and Algebra 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
 July 26th, 2016, 08:41 PM #5 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 $\displaystyle \prod_{i=1}^{10} log_2(2^i) = \prod_{i=1}^{10} \{ilog_2(2)\} = \prod_{i=1}^{10} (i * 1) = \prod_{i=1}^{10} i = what?$ 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.
 July 27th, 2016, 03:17 AM #6 Newbie   Joined: Jun 2016 From: New York Posts: 14 Thanks: 0 Math Focus: Calculus and Algebra Now what about the productmation (I'm not even sure that's what you call it, of B when the log base changes?
 July 27th, 2016, 04:16 AM #7 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 $\log_2(4) \cdot \log_4(6) \cdot \log_6(8 ) \cdot ... \cdot \log_{14}(16)$ 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?
 July 27th, 2016, 07:11 AM #8 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 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. Thanks from skeeter
July 27th, 2016, 06:03 PM   #9
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Quote:
 Originally Posted by JeffM1 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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