@@nasrullahhusnan2289 Yes. And no…You can solve it by luckily just guessing 1 and plugging it in (like I did). Or you can actually solve for / factor out the value(s) of X just using math.
@@skateordiee: Why not? [(2³)^x]+(2^x)=10 (2^x)³+(2^x)-10=0 [(2^x)³-8]+[(2^x)-2]=0 As 8=2³, factor out (2^x)-2: [(2^x)-2][{(2^x)²+2(2^x)+4}+1]=0 --> 2^x=2 hence x=1 (2^x)²+2(2^x)+5=0 It's a quadratic equation in 2^x 2^x=½[-4±sqrt(-16)] =2(1±i) We reject it as 2^x>0 Anything wrong?
Se y=2^x, allora y^3+Y=10 => y(1+y^2)=10 . 10 =2*5 y=2; y^2+1=5 => 2^x=2 e 2^(2x)=5-1=4 ovvero a 2^x=2 la cui soluzione è x=1
I like your way better, but somehow I just knew it was 1 lol
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Do you agree that the problem is problem for math olympiad?
Do you agree that the problem is math olympiad problem?
@@nasrullahhusnan2289 Yes. And no…You can solve it by luckily just guessing 1 and plugging it in (like I did). Or you can actually solve for / factor out the value(s) of X just using math.
@@skateordiee: Why not?
[(2³)^x]+(2^x)=10
(2^x)³+(2^x)-10=0
[(2^x)³-8]+[(2^x)-2]=0
As 8=2³, factor out (2^x)-2:
[(2^x)-2][{(2^x)²+2(2^x)+4}+1]=0
--> 2^x=2 hence x=1
(2^x)²+2(2^x)+5=0
It's a quadratic equation in 2^x
2^x=½[-4±sqrt(-16)]
=2(1±i)
We reject it as 2^x>0
Anything wrong?
1.