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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
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Quote:
 Originally Posted by greg1313 You raise both sides to the power of 2
That would be "squaring". I believe you mean " raise 2 to the power of each side" or something similar.

December 22nd, 2015, 04:09 PM   #13
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Quote:
 Originally Posted by v8archie That would be "squaring". I believe you mean " raise 2 to the power of each side" or something similar.
Oops. That's correct.

December 22nd, 2015, 07:48 PM   #14
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Quote:
 Originally Posted by greg1313 No, you do not "divide ... by $\log_2$". You raise 2 to the power of both sides, or use the relation given above (by skeeter).
Ok. But can we say 'cancel the $\log_2$ of both sides?'

 December 22nd, 2015, 07:52 PM #15 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,881 Thanks: 1088 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: 724 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: 7,502 Thanks: 2511 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
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Quote:
 Originally Posted by Chikis Ok! I see now. At the end I get $x=\sqrt{2}$

it's good, it's very good ...

but the other sees you?

December 23rd, 2015, 06:34 AM   #20
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Quote:
 Originally Posted by v8archie 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 $0=\pi$?
I know that this is rhetorical, but you could actually cancel those if you were so inclined. To cancel the plus while keeping the first addend equal, you could get 1 - 4 = 1 - 1 - 3 = -3, or 2 - 3 = 2 - 3 = -1.

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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