A Nice Exponential Equation

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[@jim2376]

Hint: 3 + 4 - 6 = 1 and 1 + 1 - 1 = 1. So x = 1 or 0

[@TedHopp]

That shows two solutions, but doesn't show that those are the only ones.

[@jim2376]

@@TedHopp What are the other solutions?

[@TedHopp]

@@jim2376 There are no other solutions. My point is that just noticing two solutions doesn't establish that there are no others. It's not like with polynomials where the fundamental theorem of algebra tells you ahead of time exactly how many solutions there are. With exponential equations, some additional work is needed.

[@jim2376]

@@TedHopp Exactly! That's why my point was 1 or 0. Thank you.

[@TedHopp]

@@jim2376 Let me put it another way. The conclusion, "So x = 1 or 0" is ambiguous. It can be understood as a partial list ("So the solution set includes 1 and 0 (and possibly other values)"). If that's what you meant, then ignore everything I'm saying. But it can also be understood as exhaustive ("So the only solutions are 1 and 0"), which is how I read it. If that's what you meant, then, while the conclusion happens to be correct, it does not follow logically from the hint. The hint itself does not rule out the existence of other solutions; that must be done by other means.

[@pc4shglitch746]

7:08 Maybe I'm wrong but in the quadratic formula you put 4(a+c) instead of 4ac

[@michaeledwards2251]

The quadratic formulas most typical forms are
(a) x = ( b + - root( b^2 - 4ac))/2a
(b) x = ((b/2a) + - root( (b/2a)^2 - (c/a) )
I'm not familiar with any form using 4(a+c)

[@TedHopp]

No, he used 4ac like he was supposed to. In this case, "c" was the quantity (b-1).

[@pc4shglitch746]

I understood now, thanks​@@TedHopp