Like most of his videos, lots of unnecessary babbling, with too many tangents. This ends up not being an algebra problem, just trial and error. His trial and error is more complicated than using original equation.
At exactly 16 minutes into the video, recognize that 65 is equal to 13 x 5. This simplifies the calculation to arrive at a=9. Set a+b=13, a-b=5. Therefore, a=9, b=4.
He needs to show firstly that exact methods are inadequate and then transition to successive approximation. By ignoring of avoiding the subject he is ?? IDK.
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.
Tecnica dell'avvicinamento graduale: 3 * - 2 * = 65 ... dove * = m, a = 3 * , b= 2 * --> ci sono solo 3 possibilità, 1 per ogni diverso valore di * : (a - b) < 65 , (a - b) = 65, (a - b) > 65 si procede per tentativi: 1) * = 3 --> 27 - 8 = 19 19 243 - 32 = 211 211>> 65 3) * = 4 --> 81 - 16 = 65 65 = 65 confermato, m = 4 (Marco Alby)
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
You have to find x so that: g(x)=3^m=65+2^m If m>=0 then both g(m) and f(m) are monotonocies funciona. But g(m) grows faster than f(m) As f(0)=66>1;g(m)=f(m) for some m* and after that as g(m) grows faster than f(m), g(m)>f(m) for m>x*. Só m* is unique for m>=0. for m
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.
The simplest solution is this: First m = 4 is a solution, by trial and error. Then 3^m - 2^m = 3^m . (1 - (2/3)^m) and this is obviously an increasing function. So for m > 4, 3^m - 2^m is > 65. So there is no other solution.
dunno how anyone else does it, but I generally start with a ballpark guess. 3^m >65, 3^4=81, 81-65=16= 2^4, problem solved. Any other possibilities? Negative? Fractional? Imaginary? Don't think so, but proofs would be interesting.
OK, that's relatively easy to work out by trial and error, so how about solving this equation where the right hand side is 536.784 instead of 65. I know what m is because I chose a random figure to 3 decimal places and plugged it into the equation, but I'd like to see how you or anyone else solves this. I don't think I can.
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.
Twice at the beginning of your video you said 3m - 2m instead of "3 to the m minus 2 to the m" Secondly , at no point di you use a method other than "just trial and error" method, what if the equation had been 3^m - 2^m = 64 ?
Why was negative 4 ignored? Okay, negative 5 as well? I suppose it'll be because you "deduced or induced" with the simple test what the solution must be, so you could just pick the simplest path that reached paydirt. But I still have an itch.
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
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.
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 !!!
A little rant: It’s irksome that a lot of the comments read as; “It’s all about me, I’ve already learnt this stuff so you shouldn’t bother taking the time to try to help other people learn stuff I already know.” “I didn’t know something then I gained the knowledge and now it makes me feel good about myself to tell people, who haven’t gained the knowledge, how easy it is for me”, Have some humility.
You are right on. The ones making the negative remarks are, apparently, so knowledgeable in math; then why are they watching this? Let those who want to learn, watch and learn.
Hey there, Mr. TH-cam Math Man! I've been really intrigued by this problem! Please note that for the duration of this post, I'm using x and y in place of a and b. As far as I can tell, you can only solve for the value of m using the difference of squares formula, x^2-y^2=(x+y)(x-y), when m is both a product of 2 and is greater than or equal to 2. In other words, an even number greater than or equal to 2. Thus, m can have a value of 2, 4, 6, 8, 10, 12, and so on. This is a direct result of the conditions established at the beginning of the problem by making x=3^(m/2) and y=2^(m/2). Therefore, m itself must always be a product of 2 in order to yield a whole number. For example, as seen in this problem, if m=4, then m/2=4/2=2. However, if m=3, you end up with the ratio m/2=3/2, which obviously doesn't yield a whole number when divided. Anyway, the value on the right side of the equation will need to change depending on the value of m so that the equation remains true. If m=4, we get 3^m-2^m=65, the equation used here. If m=2, the equation becomes 3^m-2^m=5. However, if m has a value that isn't a product of 2, you can't use the difference of squares formula to solve for it. Regardless of which pair of factors you use, you will end up with two numbers that can't be broken down into an exponent of either base 2 or 3. Even if you try using logarithms, you won't obtain the correct value for m. For example, when m=3, the equation becomes 3^m-2^m=19, which can't be solved using the difference of squares formula. However, it can be solved using the difference of cubes formula, where x^3-y^3=(x-y)(x^2+xy+y^2). In this case, the equation takes the form of x^3-y^3=19, where x=3^(m/3) and y=2^(m/3). However, just as using the difference of squares formula requires that m be a product of 2, so too does using the difference of cubes formula requires that m be a product of 3. In other words, m is both a product of 3 and is greater than or equal to 3. Thus, m can have a value of 3, 6, 9, 12, 15, 18, and so on. Once again, this is a direct result of the conditions established at the beginning of the problem by making x=3^(m/3) and y=2^(m/3). Therefore, m itself must always be a product of 3 in order to yield a whole number. For example, if m=3, then m/3=3/3=1. However, if m=4, you end up with the ratio m/3=4/3, which obviously doesn't yield a whole number when divided. Interestingly enough, just as using the difference of squares formula gives one set of (x,y) values, using the difference of cubes formula gives two sets of (x,y) values. In the case of x^3-y^3=19, where x=3^(m/3) and y=2^(m/3), one set of (x,y) values will be positive while the other will be negative. Obviously, only the positive set of (x,y) values will yield the correct solution. If m=6, the equation becomes 3^m-2^m=665. In this case, the value of m can be obtained by using either the difference of squares formula or the difference of cubes formula because 6 is evenly divisible by both 2 and 3. m/2=6/2=3 and m/3=6/3=2. For this reason, both formulas can be used when m has a value of 6, 12, 18, 24, 30, 36, and so on. This further implies that if m=4, you could also use x^4-y^4. If m=5, you would use x^5-y^5. If m=6, you could also use x^6-y^6, and so on. Obviously, using this approach will make your equations far more complex and not very practical each time the value of m increases. This also implies that using the difference of either squares or cubes won't solve for values of m which aren't evenly divisible by either 2 or 3, such as 5, 7, 11, 13, 17, 19, and so on. Also, things get kind of weird when m equals either 0 or 1. To sum up, use the difference of squares formula when m is a product of 2. Use the difference of cubes formula when m is a product of 3. And use either formula when m is a product of 6. Anyway, getting back to the original problem, when m is a product of 2, is greater than or equal to 2, and thus can be solved for using the difference of squares, only the middle pair of factors of the number on the right side of the equation will yield the correct solution. For example, if we take the equation presented in the video, 65 has four factors, which are 1, 5, 13, and 65. Only by using the two middle factors, 5 and 13, do we obtain the correct solution. Similarly, when m=8, the equation becomes 3^m-2^m=6305. 6305 has eight factors, which are 1, 5, 13, 65, 97, 485, 1261, and 6305. Once again, only by using the two middle factors, 65 and 97, do we obtain the correct solution. However, when m is a factor of 3, is greater than or equal to 3, and thus can be solved for using the difference of cubes, then it's possible for another pair of factors of the number on the right side of the equation to yield the correct solution instead of the middle two. This is especially true when m has a value of either 6 or 12. In any case, it's a process of rejection and elimination to find which set of factors gives the correct value of m. Finding the factors of any number is easy when using a website such as Number Empire. In addition to algebra, graphing is another ideal method for finding the value of m. If we use the equation presented here, 3^m-2^m=65, let m=x, then move everything to one side and make it equal to 0 or y, we get y=3^x-2^x-65. This produces a logarithmic graph with (x,y) intercepts at (4,0) and (0,-65). In other words, the graph intersects the y axis at -65 and the x axis at 4. If x=m, then 4 is our answer. For equations that prove difficult to solve using algebra, such as when m equals 0 or 1, equations involving the sum of squares, such as 2^m+3^m=13 when m=2, or when m is a value indivisible by either 2 or 3, such as when m equals 5, 7, or 11, then graphing is a viable, convenient, and useful alternative to find the value of m. Anyway, I would love to hear your thoughts on this! Or just graph it for fun! Either way, thanks so much for the content you post here on TH-cam! I find your discussions really entertaining! Keep up the good work, Mr. TH-cam Math Man! Grá ó Colwyn 🌝🦉🌲
sorry but you keep repeating yourself too often and prolonging your explanations .as a result one loses track , this should not have taken more than 6- 7 minutes. if someone doesn’t know 2 to the power 4 means 2x2x2x2 , should not even be trying to do this equation?!?!
I found a crazy result of (a+b)=65 and (a-b)=1. The result is two values, one value for 3^m and another for 2^m The value for 3^m is m=2+2log 11{base3} and the value for 2^m is m=10, I checked them and they work, of course you’re supposed to have only one value for m across the whole equation, so that’s weird.
Man!!!!! This is unwatchable,,,,,, so much extra extraneous nonsense information, instead of just getting to the F-ing point. This method of teaching is what made school difficult for some..
This exact question has been done so many times before, why did you post this? You did exactly what the others posted with the exact amount of assumptions ie. "m" is a whole number. You could have at least tried factors of 1 and 65 and shown they did not work. At least it would have been slightly different from the other posts solving this exact same question. Or use the statement when factoring 65, we need prime factors. Just lazy content/math I think.
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)??
Like most of his videos, lots of unnecessary babbling, with too many tangents. This ends up not being an algebra problem, just trial and error. His trial and error is more complicated than using original equation.
M is 4
At exactly 16 minutes into the video, recognize that 65 is equal to 13 x 5. This simplifies the calculation to arrive at a=9. Set a+b=13, a-b=5. Therefore, a=9, b=4.
All you did was basically solve it via trial and error in a complicated way. Kind of a dud here
Not true, he ended up solving it analytically.
He needs to show firstly that exact methods are inadequate and then transition to successive approximation. By ignoring of avoiding the subject he is ?? IDK.
thanks for the lesson
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.👍
How would you solve such problem problems when m is an odd number?
by inspection, 3^4 - 2^4 = 81 - 16 = 65
Tecnica dell'avvicinamento graduale: 3 * - 2 * = 65 ... dove * = m, a = 3 * , b= 2 * --> ci sono solo 3 possibilità, 1 per ogni diverso valore di * : (a - b) < 65 , (a - b) = 65, (a - b) > 65
si procede per tentativi: 1) * = 3 --> 27 - 8 = 19 19 243 - 32 = 211 211>> 65 3) * = 4 --> 81 - 16 = 65 65 = 65 confermato, m = 4 (Marco Alby)
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
You have to find x so that:
g(x)=3^m=65+2^m
If m>=0 then both g(m) and f(m) are monotonocies funciona.
But g(m) grows faster than f(m)
As f(0)=66>1;g(m)=f(m) for some m* and after that as g(m) grows faster than f(m), g(m)>f(m) for m>x*.
Só m* is unique for m>=0.
for m
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.
Nope…that one was beyond me.
The simplest solution is this: First m = 4 is a solution, by trial and error. Then 3^m - 2^m = 3^m . (1 - (2/3)^m) and this is obviously an increasing function. So for m > 4, 3^m - 2^m is > 65.
So there is no other solution.
Through trial and error, m=4, but I was able to start with the observation that 3ᵐ > 65, and 3³- 27, while 3⁴=81, checking 4:
3⁴ - 2⁴ = 81 - 16 = 65
dunno how anyone else does it, but I generally start with a ballpark guess. 3^m >65, 3^4=81, 81-65=16= 2^4, problem solved. Any other possibilities? Negative? Fractional? Imaginary? Don't think so, but proofs would be interesting.
I was really good at Algebra but this one is beyond my understanding. Graphing seems to be the best way to solve the problem.
OK, that's relatively easy to work out by trial and error, so how about solving this equation where the right hand side is 536.784 instead of 65. I know what m is because I chose a random figure to 3 decimal places and plugged it into the equation, but I'd like to see how you or anyone else solves this. I don't think I can.
Using algebra I mean, not graphing. I could solve it by using an excel sheet and doing semi-manual iteration to home in on the answer.
Very nice solving but...was it luck? Can you find "m" in another similar exponential equation, or this was a special case?
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.
Ⓜ = 4 . 81- 16 = 16
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.
Looks way too simple for that.
Roller coaster ride! Thanks.
No one said why m=4 is the ONLY solution
Nice, living in the universe of whole numbers.
This is another arrow with the target drawn around it.
Twice at the beginning of your video you said 3m - 2m instead of "3 to the m minus 2 to the m"
Secondly , at no point di you use a method other than "just trial and error" method, what if the equation had been 3^m - 2^m = 64 ?
65 less and near 81=3^4 ie m=4
81_16=65
3^4_2^4=81_16=65
3^m - 2^m = 65
3^4 - 2^4 = 81 - 16 = 56
x = 4
Why was negative 4 ignored?
Okay, negative 5 as well?
I suppose it'll be because you "deduced or induced" with the simple test what the solution must be, so you could just pick the simplest path that reached paydirt. But I still have an itch.
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
3^m must be greater tan 65 since 3^m-2^m is = 65. Try m =4, 3^4 =81; Subtract 2^4 =16 gives 65 so m=4.
Thank you
הצבתי מיספרים עד שהיגעתי למיספר 4 .
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.
Interesting but takes too long to get to the point/solution.
He goes off on too many tangents. Very distracting for kids with ADHD.
it is easy to guess : m=4, 81 -16
Says I’m going to show you the answer in just a second then proceeds to talk about nothing for over a minute……..
Took less than 30 seconds. 3^4-2^4=65' m=4.
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 !!!
'exacerbated at this'
81 - 16 = 65 = 9² - 4² = 3² * 3² - 2² * 2² .. der Rest ist die Ausgangsaufgabe
e^x-√x/x=0
A little rant:
It’s irksome that a lot of the comments read as;
“It’s all about me, I’ve already learnt this stuff so you shouldn’t bother taking the time to try to help other people learn stuff I already know.”
“I didn’t know something then I gained the knowledge and now it makes me feel good about myself to tell people, who haven’t gained the knowledge, how easy it is for me”,
Have some humility.
You are right on. The ones making the negative remarks are, apparently, so knowledgeable in math; then why are they watching this? Let those who want to learn, watch and learn.
I help students with math. It would be helpful if you would stay on subject a little better.
You can solve this by trial and error method
He basically did it that way.
Skip the first 7 minutes.
Okay.... I guess that's how you do it.
You take tooo long. You’re a typical ‘bad’ teacher - you love to hear yourself talk. Worst teacher, I feel sorry for your students!!!
M is 4
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
4 .... 3 seconds
m= 5.
No need to solve, answer is m=4.
m= 4
4
कहना क्या चाहते हो😂
One can solve it by hit and trial method
Or by graphical method
Id dicitur m= 4,respondeo😅😅😅😅😅😅🎉🎉🎉🎉❤😂
The answer is 4. M=4.
Good dog! And hello from a fellow dog.
Hey there, Mr. TH-cam Math Man! I've been really intrigued by this problem! Please note that for the duration of this post, I'm using x and y in place of a and b.
As far as I can tell, you can only solve for the value of m using the difference of squares formula, x^2-y^2=(x+y)(x-y), when m is both a product of 2 and is greater than or equal to 2. In other words, an even number greater than or equal to 2. Thus, m can have a value of 2, 4, 6, 8, 10, 12, and so on. This is a direct result of the conditions established at the beginning of the problem by making x=3^(m/2) and y=2^(m/2). Therefore, m itself must always be a product of 2 in order to yield a whole number. For example, as seen in this problem, if m=4, then m/2=4/2=2. However, if m=3, you end up with the ratio m/2=3/2, which obviously doesn't yield a whole number when divided.
Anyway, the value on the right side of the equation will need to change depending on the value of m so that the equation remains true. If m=4, we get 3^m-2^m=65, the equation used here. If m=2, the equation becomes 3^m-2^m=5. However, if m has a value that isn't a product of 2, you can't use the difference of squares formula to solve for it. Regardless of which pair of factors you use, you will end up with two numbers that can't be broken down into an exponent of either base 2 or 3. Even if you try using logarithms, you won't obtain the correct value for m. For example, when m=3, the equation becomes 3^m-2^m=19, which can't be solved using the difference of squares formula.
However, it can be solved using the difference of cubes formula, where x^3-y^3=(x-y)(x^2+xy+y^2). In this case, the equation takes the form of x^3-y^3=19, where x=3^(m/3) and y=2^(m/3). However, just as using the difference of squares formula requires that m be a product of 2, so too does using the difference of cubes formula requires that m be a product of 3. In other words, m is both a product of 3 and is greater than or equal to 3. Thus, m can have a value of 3, 6, 9, 12, 15, 18, and so on. Once again, this is a direct result of the conditions established at the beginning of the problem by making x=3^(m/3) and y=2^(m/3). Therefore, m itself must always be a product of 3 in order to yield a whole number. For example, if m=3, then m/3=3/3=1. However, if m=4, you end up with the ratio m/3=4/3, which obviously doesn't yield a whole number when divided.
Interestingly enough, just as using the difference of squares formula gives one set of (x,y) values, using the difference of cubes formula gives two sets of (x,y) values. In the case of x^3-y^3=19, where x=3^(m/3) and y=2^(m/3), one set of (x,y) values will be positive while the other will be negative. Obviously, only the positive set of (x,y) values will yield the correct solution.
If m=6, the equation becomes 3^m-2^m=665. In this case, the value of m can be obtained by using either the difference of squares formula or the difference of cubes formula because 6 is evenly divisible by both 2 and 3. m/2=6/2=3 and m/3=6/3=2. For this reason, both formulas can be used when m has a value of 6, 12, 18, 24, 30, 36, and so on. This further implies that if m=4, you could also use x^4-y^4. If m=5, you would use x^5-y^5. If m=6, you could also use x^6-y^6, and so on. Obviously, using this approach will make your equations far more complex and not very practical each time the value of m increases. This also implies that using the difference of either squares or cubes won't solve for values of m which aren't evenly divisible by either 2 or 3, such as 5, 7, 11, 13, 17, 19, and so on. Also, things get kind of weird when m equals either 0 or 1.
To sum up, use the difference of squares formula when m is a product of 2. Use the difference of cubes formula when m is a product of 3. And use either formula when m is a product of 6.
Anyway, getting back to the original problem, when m is a product of 2, is greater than or equal to 2, and thus can be solved for using the difference of squares, only the middle pair of factors of the number on the right side of the equation will yield the correct solution. For example, if we take the equation presented in the video, 65 has four factors, which are 1, 5, 13, and 65. Only by using the two middle factors, 5 and 13, do we obtain the correct solution. Similarly, when m=8, the equation becomes 3^m-2^m=6305. 6305 has eight factors, which are 1, 5, 13, 65, 97, 485, 1261, and 6305. Once again, only by using the two middle factors, 65 and 97, do we obtain the correct solution. However, when m is a factor of 3, is greater than or equal to 3, and thus can be solved for using the difference of cubes, then it's possible for another pair of factors of the number on the right side of the equation to yield the correct solution instead of the middle two. This is especially true when m has a value of either 6 or 12. In any case, it's a process of rejection and elimination to find which set of factors gives the correct value of m. Finding the factors of any number is easy when using a website such as Number Empire.
In addition to algebra, graphing is another ideal method for finding the value of m. If we use the equation presented here, 3^m-2^m=65, let m=x, then move everything to one side and make it equal to 0 or y, we get y=3^x-2^x-65. This produces a logarithmic graph with (x,y) intercepts at (4,0) and (0,-65). In other words, the graph intersects the y axis at -65 and the x axis at 4. If x=m, then 4 is our answer.
For equations that prove difficult to solve using algebra, such as when m equals 0 or 1, equations involving the sum of squares, such as 2^m+3^m=13 when m=2, or when m is a value indivisible by either 2 or 3, such as when m equals 5, 7, or 11, then graphing is a viable, convenient, and useful alternative to find the value of m.
Anyway, I would love to hear your thoughts on this! Or just graph it for fun! Either way, thanks so much for the content you post here on TH-cam! I find your discussions really entertaining! Keep up the good work, Mr. TH-cam Math Man!
Grá ó Colwyn 🌝🦉🌲
sorry but you keep repeating yourself too often and prolonging your explanations .as a result one loses track , this should not have taken more than 6- 7 minutes.
if someone doesn’t know 2 to the power 4 means 2x2x2x2 , should not even be trying to do this equation?!?!
3^4-2^4 = 65
m = 4 no calculator.
Or ln65/ln3 = 4.0 = m
Perhaps not so academic but it works.
I found a crazy result of (a+b)=65 and (a-b)=1. The result is two values, one value for 3^m and another for 2^m
The value for 3^m is m=2+2log 11{base3} and the value for 2^m is m=10, I checked them and they work, of course you’re supposed to have only one value for m across the whole equation, so that’s weird.
Why not take an hour? Add a few more smiley and sad faces and blah blah blah
It is just one minute to solve this question,
This vdo clip is 21 minutes,
..... What the babbling.....
Too much repetition
Teacher, you are talking too much...make it shorter...Hi from Croatia
M=4. 🤗🤗 Way too easy. Me about 6 seconds.
You love to talk talk talk. Get to the point. Not everyone has time on their hands as you 👎
The fact that this took almost 22 minutes to explain is why I hated math.
Just put 4 in place of m
Man!!!!! This is unwatchable,,,,,, so much extra extraneous nonsense information, instead of just getting to the F-ing point. This method of teaching is what made school difficult for some..
Talk too much....❤
This exact question has been done so many times before, why did you post this? You did exactly what the others posted with the exact amount of assumptions ie. "m" is a whole number. You could have at least tried factors of 1 and 65 and shown they did not work. At least it would have been slightly different from the other posts solving this exact same question. Or use the statement when factoring 65, we need prime factors. Just lazy content/math I think.
I can look at it and tell u 4. But do you want all the imaginary solutions too?
Yes.
m=4 is the answer
3⁴ -2⁴= 81-16=65
4
Nice, living in the universe of whole numbers.
M=4
4
m=4
m=4
m = 4
m=4
m = 4
m=4
M=4