Thank you, Professor, another great example. I love that you take nothing for granted and you always pepper your lessons with all the required steps and identities.
For complex m solution iz NOT undefined but complex value(s). Complex numbers could be writen in polar and exponential form. Any complex number Z=a+i×b may be rewriten as Z=R×(cos(teta)+i×sin(teta)), where R=sqrt(a^2+b^2) and teta=arctan(b/a). This could be further rewriten as Z=R×e^(i×teta). Note that due to periodicity of trigonometric function there are indefinitely many values of teta that are arctan(b/a) in form p+nPI where p is principal value and n is whole number, and PI=3.14.... Then you can find natural logarithm of exponential function. ln(2^x)=ln(R×e^(i×teta)) ln2×x = ln(R)+i×teta X= (lnR+i×teta)/ln2 In your case a1,2=2 b1=2×sqrt(2) and b2=-2×sqrt(2). R is sqrt(4+8) = sqrt(12)= 2sqrt(3) in both cases. Principal values for teta are arctan(sqrt(2)) , arctan(-sqrt(2))=-1×arctan(sqrt(2)).
I agree with @GoranJakupovic - I think that in presentations that go into such great detail it is important not to make erroneous "throw-away" remarks - it undermines the main point of giving rigorous solutions. Indeed it is obvious that, by the method in the video, the complex cases are x =(log[-2+/-2*i*sqrt[2]] +2*pi*n*i)/log[2]. Admittedly simplifying that takes some time and introducing log[-1] and log[i] but, by construction, the complex expressions for m satisfy the equation.
@@tassiedevil2200,exactly, complex solution could be discarded only if problem stated something like "find all real solutions of ...". Even then one should show that complex m does not give real x. I beleive that main point of video is to demonstrate how to use substition to convert exponential equation to cubic, and how to solve cubic equation by grouping (instead using something like Cardano formula for cubic equation), however by disregarding complex solution this gives incomplete solution set. Even if problem was defined to find real solutions, there should be info that complex solutions exist, but that they are out of scope for video.
Thank you, Professor, another great example. I love that you take nothing for granted and you always pepper your lessons with all the required steps and identities.
My pleasure! I appreciate that! Excellent!💕🙏💖👌💪✅
For complex m solution iz NOT undefined but complex value(s). Complex numbers could be writen in polar and exponential form. Any complex number Z=a+i×b may be rewriten as Z=R×(cos(teta)+i×sin(teta)), where R=sqrt(a^2+b^2) and teta=arctan(b/a). This could be further rewriten as Z=R×e^(i×teta). Note that due to periodicity of trigonometric function there are indefinitely many values of teta that are arctan(b/a) in form p+nPI where p is principal value and n is whole number, and PI=3.14.... Then you can find natural logarithm of exponential function. ln(2^x)=ln(R×e^(i×teta))
ln2×x = ln(R)+i×teta
X= (lnR+i×teta)/ln2
In your case a1,2=2 b1=2×sqrt(2) and b2=-2×sqrt(2).
R is sqrt(4+8) = sqrt(12)= 2sqrt(3) in both cases.
Principal values for teta are arctan(sqrt(2)) , arctan(-sqrt(2))=-1×arctan(sqrt(2)).
Thanks for your in-depth analysis of the complex solution! 💯🤩🙏✅
I agree with @GoranJakupovic - I think that in presentations that go into such great detail it is important not to make erroneous "throw-away" remarks - it undermines the main point of giving rigorous solutions. Indeed it is obvious that, by the method in the video, the complex cases are x =(log[-2+/-2*i*sqrt[2]] +2*pi*n*i)/log[2]. Admittedly simplifying that takes some time and introducing log[-1] and log[i] but, by construction, the complex expressions for m satisfy the equation.
@@tassiedevil2200,exactly, complex solution could be discarded only if problem stated something like "find all real solutions of ...". Even then one should show that complex m does not give real x.
I beleive that main point of video is to demonstrate how to use substition to convert exponential equation to cubic, and how to solve cubic equation by grouping (instead using something like Cardano formula for cubic equation), however by disregarding complex solution this gives incomplete solution set. Even if problem was defined to find real solutions, there should be info that complex solutions exist, but that they are out of scope for video.
let u=2^x , u^3+u^2+/-u-36=0 , (u-3)(u^2+4u+12)=0 , 2^x=3 , x=log3/log2 , test , 27+9=36 , OK ,
1 -3
4 -12
12 -36
Решаем методом устного счета.
2^х=y
y^2(y+1)=9×4
y=3
x=log3
Good luck!
x= log_2(3)
А где твоё решение ?
2(3) ?
@@davidseed2939