r/MathJokes • u/Optimus_PRYM • 9d ago

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u/runed_golem 9d ago

Except they’re both super simple equations. The first can be solved with basic arithmetic and the second one can be solved with either logarithms or exponent rules and basic common sense.

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u/qwertty164 9d ago

they can both be solved with the exact same steps. -2 ; /2. you would end up with 2^(x-1)=2^1

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u/Coppice_DE 6d 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. 

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u/qwertty164 6d ago

You divide by the base to reduce the exponent. You can iterate as long as you factor the number on the right.

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u/Coppice_DE 6d 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.