(3 to the m) - (2 to the m) = 65 m=? MOST won’t FIGURE OUT how to solve!

Рет қаралды 59,480

Күн бұрын

Пікірлер: 97

@tomtke7351 Ай бұрын

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)??

@robertjohnson4401 Ай бұрын

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.

@richardmullins44 8 күн бұрын

by inspection, 3^4 - 2^4 = 81 - 16 = 65

@russelllomando8460 Ай бұрын

thanks for the lesson

@marcoalby7499 13 күн бұрын

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)

@dericcadora2801 15 күн бұрын

All you did was basically solve it via trial and error in a complicated way. Kind of a dud here

@jimmacaulay844 3 күн бұрын

Not true, he ended up solving it analytically.

@rfarrelldic Ай бұрын

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

@roccov11 7 күн бұрын

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.

@roccov11 7 күн бұрын

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.

@kajamix 8 күн бұрын

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.

@gregorynelson1568 4 күн бұрын

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

@pedrojose392 8 күн бұрын

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

@armchairtin-kicker503 Ай бұрын

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.

@Dr_piFrog Ай бұрын

This is the proper solution, not the round-robin third treatment presented in the video.👍

@jimselvy6157 15 күн бұрын

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.

@arthurvandeth3846 8 күн бұрын

M is 4

@Kleermaker1000 Ай бұрын

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.

@gspaulsson 11 күн бұрын

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.

@andrespataky8410 14 сағат бұрын

Very nice solving but...was it luck? Can you find "m" in another similar exponential equation, or this was a special case?

@fernandocagarrinho3980 Ай бұрын

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.

@reinymichel 7 күн бұрын

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 ?

@conniebartley4226 13 күн бұрын

I was really good at Algebra but this one is beyond my understanding. Graphing seems to be the best way to solve the problem.

@abdulhusseinalsultani9222 12 күн бұрын

65 less and near 81=3^4 ie m=4 81_16=65 3^4_2^4=81_16=65

@Bill_Woo 11 күн бұрын

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.

@cyruschang1904 Ай бұрын

3^m - 2^m = 65 3^4 - 2^4 = 81 - 16 = 56 x = 4

@terry_willis Ай бұрын

This problem should be in the Math Olympiad.

@AINeverDies Ай бұрын

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.

@CrYou575 4 күн бұрын

Looks way too simple for that.

@John-dp8oh Күн бұрын

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.

@jaggisaram4914 3 күн бұрын

Ⓜ = 4 . 81- 16 = 16

@valentin5403 9 күн бұрын

No one said why m=4 is the ONLY solution

@johnlakin5895 Ай бұрын

Nope…that one was beyond me.

@bobwilson3980 12 күн бұрын

This is another arrow with the target drawn around it.

@Rich.Staples Ай бұрын

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

@thomashawkinson7017 Ай бұрын

Roller coaster ride! Thanks.

@DennisMcFall Ай бұрын

Interesting but takes too long to get to the point/solution.

@jtinalexandria 23 күн бұрын

He goes off on too many tangents. Very distracting for kids with ADHD.

@wilgoster 24 күн бұрын

Nice, living in the universe of whole numbers.

@wes9627 Ай бұрын

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.

@rarocon Ай бұрын

81 - 16 = 65 = 9² - 4² = 3² * 3² - 2² * 2² .. der Rest ist die Ausgangsaufgabe

@mehdizangiabadi-iw6tn Ай бұрын

e^x-√x/x=0

@vijaybhatia2686 Ай бұрын

What if the problem was 3*m + 2*m = 97 ?

@AINeverDies Ай бұрын

It's 110% easier just combine like terms, and it becomes obvious: 3m+2m=97 5m=97 m=97/5 m=19.4

@tanveernaeem1157 15 күн бұрын

Took less than 30 seconds. 3^4-2^4=65' m=4.

@josephlaura7387 Ай бұрын

Thank you

@panlomito Ай бұрын

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 !!!

@melissalevine2373 13 күн бұрын

I help students with math. It would be helpful if you would stay on subject a little better.

@syedfazlullah6459 Ай бұрын

You can solve this by trial and error method

@CrYou575 4 күн бұрын

He basically did it that way.

@alexanderizhaki1560 10 күн бұрын

it is easy to guess : m=4, 81 -16

@sammymcclelland1071 6 сағат бұрын

M is 4

@aajjuudyo 6 күн бұрын

m= 5.

@mauriziograndi1750 Ай бұрын

3^4-2^4 = 65 m = 4 no calculator. Or ln65/ln3 = 4.0 = m Perhaps not so academic but it works.

@notreallydavid 29 күн бұрын

'exacerbated at this'

@MrGfmassot Ай бұрын

Okay.... I guess that's how you do it.

@ferrantepallas 15 күн бұрын

4 .... 3 seconds

@AllDogsAreGoodDogs Ай бұрын

The answer is 4. M=4.

@discobear5752 Ай бұрын

Good dog! And hello from a fellow dog.

@colwynbrennen 11 күн бұрын

Hey there, Mr. KZbin 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 KZbin! I find your discussions really entertaining! Keep up the good work, Mr. KZbin Math Man! Grá ó Colwyn 🌝🦉🌲

@Majan-v8K Ай бұрын

m= 4

@sandeepsantoriya822 7 күн бұрын

कहना क्या चाहते हो😂 One can solve it by hit and trial method Or by graphical method

@meshulamoren8633 12 күн бұрын

הצבתי מיספרים עד שהיגעתי למיספר 4 .

@roshanjain8374 9 күн бұрын

No need to solve, answer is m=4.

@aonghusobroin2959 Ай бұрын

@1234larry1 Ай бұрын

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.

@Neil-e4n 29 күн бұрын

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.

@allenshallbetter 21 күн бұрын

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.

@boonraypipatchol7295 Ай бұрын

It is just one minute to solve this question, This vdo clip is 21 minutes, ..... What the babbling.....

@syedmdabid7191 9 күн бұрын

Id dicitur m= 4,respondeo😅😅😅😅😅😅🎉🎉🎉🎉❤😂

@holoparandeh4467 Ай бұрын

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?!?!

@prpa9210 15 күн бұрын

You take tooo long. You’re a typical ‘bad’ teacher - you love to hear yourself talk. Worst teacher, I feel sorry for your students!!!

@KHALEDMOFTAH-f8f 8 күн бұрын

Too much repetition

@syedfazlullah6459 Ай бұрын

Just put 4 in place of m

@nikolaili1003 Күн бұрын

Teacher, you are talking too much...make it shorter...Hi from Croatia

@jas2819 Ай бұрын

M=4. 🤗🤗 Way too easy. Me about 6 seconds.

@billmorgan2883 20 күн бұрын

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.

@BMT3691 8 күн бұрын

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..