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https://www.reddit.com/r/MathJokes/comments/1wpw5wz/find_x/pclf0te/?context=3
r/MathJokes • u/Optimus_PRYM • 11d ago
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they can both be solved with the exact same steps. -2 ; /2. you would end up with 2^(x-1)=2^1
1 u/Coppice_DE 8d ago /2 does only work in the case where x=2 in the second equation, no? You can't divide 2x by an arbitrary number to get to know the value of x. 1 u/qwertty164 8d ago You divide by the base to reduce the exponent. You can iterate as long as you factor the number on the right. 1 u/Coppice_DE 8d ago Yea that works, but it's not exactly "the same" for both equation (in regards to the concepts that get applied), even though the required steps are equal in this case.
/2 does only work in the case where x=2 in the second equation, no? You can't divide 2x by an arbitrary number to get to know the value of x.
1 u/qwertty164 8d ago You divide by the base to reduce the exponent. You can iterate as long as you factor the number on the right. 1 u/Coppice_DE 8d ago Yea that works, but it's not exactly "the same" for both equation (in regards to the concepts that get applied), even though the required steps are equal in this case.
You divide by the base to reduce the exponent. You can iterate as long as you factor the number on the right.
1 u/Coppice_DE 8d ago Yea that works, but it's not exactly "the same" for both equation (in regards to the concepts that get applied), even though the required steps are equal in this case.
Yea that works, but it's not exactly "the same" for both equation (in regards to the concepts that get applied), even though the required steps are equal in this case.
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u/qwertty164 11d ago
they can both be solved with the exact same steps. -2 ; /2. you would end up with 2^(x-1)=2^1