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 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???
Attached Images image.jpg (12.4 KB, 15 views) 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 $log_2(4) = \dfrac{log_2(4)}{1} = \dfrac{log_2(4)}{log_2(2)}.$ $\displaystyle \prod_{i=1}^7 log_{2i}\{2(i+1)\} = \prod_{i=1}^7 \dfrac{log_2\{2(i+1)\}}{log_2(2i)}= \dfrac\prod_{i=1}^7 log_2 \{2(i + 1)\}}\prod_{i=1}^7 log_{2} (2i)}$ $\dfrac\prod_{i=2}^8log_2(2i)}\prod_{i=1}^7 log_2(2i)} = \dfrac{log_2(8 * 2) * \cancel\prod_{i=2}^7 log_2(2i)}}{log_2(2 * 1) * \cancel\prod_{i=2}^7 log_2(2i)}} = \dfrac{log_2(16)}{log_2(2)} = 4$ In this case, skeeter's notation is more intuitive. Thanks from skeeter July 27th, 2016, 06:03 PM   #9
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 Originally Posted by JeffM1 This is skeeter's answer in the pi notation. Note that $log_2(4) = \dfrac{log_2(4)}{1} = \dfrac{log_2(4)}{log_2(2)}.$ $\displaystyle \prod_{i=1}^7 log_{2i}\{2(i+1)\} = \prod_{i=1}^7 \dfrac{log_2\{2(i+1)\}}{log_2(2i)}= \dfrac\prod_{i=1}^7 log_2 \{2(i + 1)\}}\prod_{i=1}^7 log_{2} (2i)}$ $\dfrac\prod_{i=2}^8log_2(2i)}\prod_{i=1}^7 log_2(2i)} = \dfrac{log_2(8 * 2) * \cancel\prod_{i=2}^7 log_2(2i)}}{log_2(2 * 1) * \cancel\prod_{i=2}^7 log_2(2i)}} = \dfrac{log_2(16)}{log_2(2)} = 4$ In this case, skeeter's notation is more intuitive.
I'm just an "intuitive" kind of guy ...  Tags dis, logarithms Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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