How to simplify and Solve the Variable ‘M’ in this Algebraic Exponential Equation? 

Yes well said 👍🏽. Thanks 🙏🏽

@@YAWTon well that is true, but finding ways to solve equations like these, you got to admit it’s pretty awesome !

@@professor_1o1 Yes I find it awesome to find ways to solve equations. However, I do not find the method that you use in the clip particularly awesome. The solution m=4 in this example is too obvious, and I would not even start to think about a strategy to solve the problem through algebraic means. But more importantly: your approach has worked in this case, but this is more or less by accident. Interestingly, you did not give a justification for factoring 65 as 5*15. Obviously, you chose the factors 5 and 13 because you knew that they would lead to the solution m=4. Try to use your method on the problem 3^m-2^m=211. Again, the solution m=5 is pretty obvious (m must be at leas 5, and 3^5-2^5=243-32=211). Applying the method from your clip, one has to find x,y such that x^2-y^2=211. But in this case, the factors ar not integer factors of 211, but irrational numbers x=9√3 and y=4√2.☹

@@professor_1o1this is my calculation: 3^m-2^m= 65, 3-2= 1, so 1=65*2m^2,1-2m^2=65, 2m^2=64, 2m=8, so m=4. Am i correct or nah?

@@Bazzmonkey nah, you can’t subtract with terms with unknown terms like that, and there are other errors as well, but not good true

He chose 2/2 to have square and 1/2 was a byproduct of this. You could do 3/3, but x^3-y^3=65 would be a hustle

@@osabossa excellent 👏🏼

@@elizandadingle9503 Yes you can use other equivalents of 1. However the greater the power is the more working for the solution.

m=1, X. m=2, X. m=3, X. m=4. VV

Before watching the video: 3^m = log(3)m = ln(m)/ln(3) 2^m = log(2)m = ln(m)/ln(2) ln(m)/ln(3) - ln(m)/ln(2) = 65 ln(m)*(1/ln(3) - 1/ln(2)) = 65 ln(m) = 65/(1/ln(3) - 1/ln(2)) m = e^(65/(1/ln(3) - 1/ln(2))) = 9.619*10^-54 Plugging in for m, I get 0. Not sure what I did wrong here.