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December 22nd, 2015, 03:36 PM  #11 
Member Joined: Dec 2015 From: Down Under Posts: 32 Thanks: 3 
I suppose technically you can call it dividing if you understand that what you are doing is dividing by log2(x)/x where x is the operand. lol Although that won't be very helpful for manipulating the equation xD Last edited by Relentless; December 22nd, 2015 at 03:38 PM. 
December 22nd, 2015, 04:06 PM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra  
December 22nd, 2015, 04:09 PM  #13 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond  
December 22nd, 2015, 07:48 PM  #14 
Senior Member Joined: Jan 2012 Posts: 725 Thanks: 7  
December 22nd, 2015, 07:52 PM  #15 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 
I suppose we could. One might prefer more accurate terminology, though.

December 22nd, 2015, 07:55 PM  #16 
Member Joined: Dec 2015 From: Down Under Posts: 32 Thanks: 3 
Well, applying the inverse function to the log would be to raise the base to the operand. I think technically that would be cancelling the log. For example, log2(x) = log2(y) becomes 2^x = 2^y, and we can see why they are equal now. I think you just have to write that log2(x) = log2(y) IMPLIES x = y Otherwise the long way you'd then take the log of both sides and end up with x*log2 = y*log2 and then you can divide lol Last edited by Relentless; December 22nd, 2015 at 08:05 PM. 
December 22nd, 2015, 08:04 PM  #17 
Senior Member Joined: Jan 2012 Posts: 725 Thanks: 7 
Yes, that is correct. Your post answers it all.

December 22nd, 2015, 09:00 PM  #18 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
Why are you so desperate to cancel notation? What if I write $1+4=2+3$ and cancel the $+$, so that $14=23$? Is that valid? Or is it nonsense? What about if I write $\sin 0 = \sin \pi$ and cancel the $\sin$ to get $0=\pi$? 
December 23rd, 2015, 06:17 AM  #19 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176  
December 23rd, 2015, 06:34 AM  #20  
Member Joined: Dec 2015 From: Down Under Posts: 32 Thanks: 3  Quote:
To cancel the sin, use inverse sin, and you will get 1/sinx = 1/siny (although it doesn't work for pi * integer). What you don't always get of course, as you mean to imply, is that the subjects are equal, even if the operations are. Last edited by Relentless; December 23rd, 2015 at 06:37 AM.  

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