Here's the last half of the solution, employing a system of linear equations. . let a = 3^(m/2) and b = 2^(m/2) a^2 - b^2 = 65 (a - b)(a+b) = 65 Factoring 65... (a-b)(a+b) = 5*13 a - b = 5 [equ. #1] a + b = 13 [equ. #2] 2a = 18 [equ. #1 plus equ. #2] a = 9 Back substituting the value 9 for variable-a... (9) = 3^(m/2) 3^2 = 3^(m/2) log(3^2) = log(3^(m/2)) 2log3 = (m/2)log3 2 = m/2 4 = m m = 4 Therefore, m is equal to 4.
solution number three is innovative but it is really trial and error again .... and if we are going to sell by trial and error error, it's much easier to just do it out right with the original equation
I didn't know where to start but you can still try some numbers by estimation. So I started with 3^4 = 81... quite nice! 81 - 65 = 16 = 2^4 and bingo !!!
How about 3^m-2^m-64=0 or 3^m-2^m-66=0 or more generally f(m)=0 where m exists and is real? Iterative solution methods such as fixed-point or Newton-Raphson iteration may be used to approximate one or more values, if they exist. When m is presumed to be a posaiive integer a good place to start, since 3^m>2^m, is m=ceiling(ln65/ln3)=4, giving 3^4-2^4=81-16=65.
This is just the same as 81 - 16 = 65 and then transform 81 into 3^m (or 3^4) and 16 into 2^m (or 2^4). This is no real algebra imho, even though m = 4.
How about proving that m has to be even for the result to be a multiple of 5. And equating prime factors of 65=5*13 to solve the difference of squares. That would teach something.
Serious question for the really smart people on this thread. What is the connection of this type of calculation to the ability to calculate reentry orbits?
why not bring the whole equation to the (m/1) power? this would eliminate the variable power over 3 and -2 and put the 65 to the (m/1) power? u'd then end up with m=0
Here's the last half of the solution, employing a system of linear equations.
.
let a = 3^(m/2) and b = 2^(m/2)
a^2 - b^2 = 65
(a - b)(a+b) = 65
Factoring 65...
(a-b)(a+b) = 5*13
a - b = 5 [equ. #1]
a + b = 13 [equ. #2]
2a = 18 [equ. #1 plus equ. #2]
a = 9
Back substituting the value 9 for variable-a...
(9) = 3^(m/2)
3^2 = 3^(m/2)
log(3^2) = log(3^(m/2))
2log3 = (m/2)log3
2 = m/2
4 = m
m = 4
Therefore, m is equal to 4.
This is the proper solution, not the round-robin third treatment presented in the video.👍
3^m - 2^m =65
stumble around
m=5 =>
3^5-2^5=?65
81×3-16×4=?65
243-64=?65
179 /=65
m=4
3^4-2^4=?65
81-16=?65
65=❤65✔️
3^m-2^m=65
÷2^m{3^m-2^m=65}
(3^m/2^m) - 1 = 65/2^m
1.5^m - 1 = 65/2^m
1.5^m - 65/2^m = 1
X^2 - Y^2 = (X + Y)(X - Y)??
solution number three is innovative but it is really trial and error again .... and if we are going to sell by trial and error error, it's much easier to just do it out right with the original equation
thanks for the lesson
I didn't know where to start but you can still try some numbers by estimation. So I started with 3^4 = 81... quite nice!
81 - 65 = 16 = 2^4 and bingo !!!
How about 3^m-2^m-64=0 or 3^m-2^m-66=0 or more generally f(m)=0 where m exists and is real? Iterative solution methods such as fixed-point or Newton-Raphson iteration may be used to approximate one or more values, if they exist. When m is presumed to be a posaiive integer a good place to start, since 3^m>2^m, is m=ceiling(ln65/ln3)=4, giving 3^4-2^4=81-16=65.
This is just the same as 81 - 16 = 65 and then transform 81 into 3^m (or 3^4) and 16 into 2^m (or 2^4). This is no real algebra imho, even though m = 4.
How about proving that m has to be even for the result to be a multiple of 5.
And equating prime factors of 65=5*13 to solve the difference of squares.
That would teach something.
Thank you
3^m - 2^m = 65
3^4 - 2^4 = 81 - 16 = 56
x = 4
Roller coaster ride! Thanks.
Serious question for the really smart people on this thread. What is the connection of this type of calculation to the ability to calculate reentry orbits?
e^x-√x/x=0
why not bring the whole equation to the (m/1) power? this would eliminate the variable power over 3 and -2 and put the 65 to the (m/1) power? u'd then end up with m=0
This problem should be in the Math Olympiad.
Well it was. I typed the problem and saw titles like "Math Olympiad (3^m) - (2^m) = 65". So probably he took the problem from that Olympiad.
81 - 16 = 65 = 9² - 4² = 3² * 3² - 2² * 2² .. der Rest ist die Ausgangsaufgabe
What if the problem was 3*m + 2*m = 97 ?
It's 110% easier just combine like terms, and it becomes obvious: 3m+2m=97 5m=97 m=97/5 m=19.4
Okay.... I guess that's how you do it.
Interesting but takes too long to get to the point/solution.
The answer is 4. M=4.
M=4. 🤗🤗 Way too easy. Me about 6 seconds.
m= 4
3^4-2^4 = 65
m = 4 no calculator.
Or ln65/ln3 = 4.0 = m
Perhaps not so academic but it works.
4
I can look at it and tell u 4. But do you want all the imaginary solutions too?
Yes.