A Nice Math Olympiad Exponential Equation | Olympiad Math | Find X
Пікірлер: 15
@PriyaBS5917Ай бұрын
When 4^x= 8,then 2^2x= 2^3. When 2x= 3, then x= 3/2. There's no need to take logarithm routes
@jim2376Ай бұрын
👍👍
@michrozАй бұрын
Just convert both parts into power of 2: 2^(2x+1)=2^4. So 2x+1=4 and x=3/2=1.5
@dentanlimАй бұрын
yup. this is much simpler
@MahakDhorey-ot3mrАй бұрын
Quadratic math 😂
@daishesАй бұрын
4^x+4^x=16 2(4^x)=16 4^x=8 (2²)^x=8 2^(2x)=2³ Exponentiation in same base with equality. The powers are the same, so we can: 2x=3 x=3/2
@Trav3l3r69.Ай бұрын
I asked ChatGPT
@primestark5475Ай бұрын
yeah why take log on such simple question. is this an olympiad question
@jcsjcs2Ай бұрын
Dude -- just take log_2 on both sides: 4^x = 8 x * log_2(4) = log_2(8) x * 2 = 3 x = 3/2.
@kablanetkablanet989Ай бұрын
Nice way it took me less than two seconds to solve this logically because only one number in the world can solve this equation Still, I enjoyed watching
@petrkinkal1509Ай бұрын
I assume you went with 4^x + 4^x = 16 so 4^x = 8 and 4 * 2 is 8 which means you need to do 4 square root of 4 (4^0.5) so it is 4^1.5. Or something like that? That is nice way to go about it but it wouldn't work if the numbers were less nice.
@kablanetkablanet989Ай бұрын
@@petrkinkal1509 First, your way is a bit similar to what I did, although my way does not have any calculations Look, I believe that mathematics is something applied and not theoretical, therefore the use of all kinds of ways disguised as logic is much more enjoyable - in my opinion, of course. In any case, the method should take into account the simple rationale any number greater than the one in front of it by one unit. Therefore, when we have an exercise (and it doesn't matter what exercise I see) we look for the difference - the difference shows the direction to think about. Hence when such an exercise came, it was not even necessary to think that the smallest number would move between 1 and 2 So the next rational number will be 1.5 By the way, it works every time and in every exercise, including extremely complicated exercises Thanks
@petrkinkal1509Ай бұрын
@@kablanetkablanet989 I somewhat doubt this statement: "By the way, it works every time and in every exercise, including extremely complicated exercises Thanks" But I can be wrong. Lets assume the answer was 43.584632 (precisely) I somewhat doubt you could get to that in reasonable amount of time (unless you are satisfied with something like 43.6) (Now to be fair you also couldn't get that in reasonable amount of time without calculator anyway.)
@kablanetkablanet989Ай бұрын
@@petrkinkal1509 What I said is that every time I used my method I was able to solve both complicated and simple exercises. So this is what is happening for me. At some point, I will try to write this method maybe it will help people. But to explain it, everyone knows the "tricks" that exist in mathematics to quickly solve multiplication exercises, squaring and more - this belongs to the same format of methods.