let y = 2^x Numerator (N)=y^3-y=y*(y^2-1)= y*(y-1)*(y+1) Denominator (D)= 3^x * (y-1) Eliminating common factor (y-1) N=y*(y+1)=2^x*( 2^x+1) D=3^x N/D=2 can be written as .. N=2*D 2^x * (2^x+1) = 2 * 3^x Now both ( 2^x+1) and 3^x are odd means both do not have factor 2 in them, so that would mean Case 1.. 2^x=2 so x=1 Case 2 2^x+1=3^x x=1 will obviously work (as per solution derived from Case 1) Any value of x > 1 or
I tried to put this equation in Wolfram Alpha for the sake of it, and it says that x = 0 is also a valid solution, but it obviously can't be, for the reason stated in the video, nice work breaking one of the most popular math tools 😁
0 is a valid solution if you take a limit as x approaches 0, because then you get closer and closer to two if x=0,000001 wolfram alpha says its basically 2 (it is 1.9999988...)
Nice work. Congrats. After some algebra the problem is reduced to (4/3)^x+(2/3)^x = 2 whose solutions are x = 0 and x = 1. The solution x = 0 can be acceptable because lim x->0 (8^x-2^x)/(6^x-3^x) = 2. In general, the solutions of a^x+b^x=c are x1=(1/ln(b/a))*ln((1/z)*Wq(z*(c^z); q = 1-1/z)) with z = ln(b/a)/ln(a) and x2=(1/ln(a/b))*ln((1/y)*Wq(y*(c^y); q = 1-1/y)) with y = ln(a/b)/ln(b). q is the parameter of the Lambert-Tsallis Wq function used. One can find more details in the paper (only two pages) "Solving the Fractional Polynomial ax^r +bx^s +c = 0 Using the Lambert-Tsallis Wq Function" that can be downloaded on Researchgate.
You must try to derive the steps more than now. Do not jump up. If You should write the reference concept or theory, it will be better than now. Thanks.
let y = 2^x
Numerator (N)=y^3-y=y*(y^2-1)= y*(y-1)*(y+1)
Denominator (D)= 3^x * (y-1)
Eliminating common factor (y-1)
N=y*(y+1)=2^x*( 2^x+1)
D=3^x
N/D=2 can be written as ..
N=2*D
2^x * (2^x+1) = 2 * 3^x
Now both ( 2^x+1) and 3^x are odd means both do not have factor 2 in them, so that would mean
Case 1..
2^x=2 so x=1
Case 2
2^x+1=3^x
x=1 will obviously work (as per solution derived from Case 1)
Any value of x > 1 or
Thank you very much dear teacher sharing with us and making everything so easy to comprehend and connect with keep going 😘👍🙏
my pleasure❣️
I tried to put this equation in Wolfram Alpha for the sake of it, and it says that x = 0 is also a valid solution, but it obviously can't be, for the reason stated in the video, nice work breaking one of the most popular math tools 😁
0 is a valid solution if you take a limit as x approaches 0, because then you get closer and closer to two
if x=0,000001 wolfram alpha says its basically 2 (it is 1.9999988...)
This confused me too
Nice work. Congrats. After some algebra the problem is reduced to (4/3)^x+(2/3)^x = 2 whose solutions are x = 0 and x = 1. The solution x = 0 can be acceptable because lim x->0 (8^x-2^x)/(6^x-3^x) = 2. In general, the solutions of a^x+b^x=c are x1=(1/ln(b/a))*ln((1/z)*Wq(z*(c^z); q = 1-1/z)) with z = ln(b/a)/ln(a) and x2=(1/ln(a/b))*ln((1/y)*Wq(y*(c^y); q = 1-1/y)) with y = ln(a/b)/ln(b). q is the parameter of the Lambert-Tsallis Wq function used. One can find more details in the paper (only two pages) "Solving the Fractional Polynomial ax^r +bx^s +c = 0 Using the Lambert-Tsallis Wq Function" that can be downloaded on Researchgate.
Спасибо.
1
Не буде коренів там...
Х=1. Моментально.
x=1
N×n !=2n
No, this is not true.
n × n = n²
n³ × n⁴ = n⁷
You must try to derive the steps more than now.
Do not jump up.
If You should write the reference concept or theory, it will be better than now.
Thanks.
这最后都是蔡来了！奇数偶数之类的？？
x=1