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Пікірлер
@Shameilia2004 4 күн бұрын
i’m sure the answer is right but the explanation for someone as novice cannot follow the steps toward the end.
@KasyannoEZMath 4 күн бұрын
Thanks for your time commenting. I appreciate it. Short videos don't explain everything. Go watch the regular length video for the full explanation on how to solve using the Lambert W Function. Click on the related video in this short video to let you or anyone who would like to learn more about solving this kind of exponential equations. The link leads to the video "Can You Solve this Exponential Equation " x⁵=4^x using the Lambert W Function detailed explanation. Happy learning!
@nathanielroosevelt1489 6 күн бұрын
Why not just start by dividing both side by 2^x, then multiply both sides by -1, then take the W, and then multiply both sides by -1 again? Am I doing something wrong because it seems a lot simpler that way
@KasyannoEZMath 4 күн бұрын
Thanks for your time commenting. I appreciate it very much. You're absolutely right. You can do that then after multiplying-1 express 2^-x = e^ln(2)^-x. Then multiply the equation by ln(2), then apply the Lambert W Function. You'll get to the alternate solution for X = -[w(-ln2)]/ln2. Happy learning!
@Watsonvan 6 күн бұрын
1/3
@jelmerterburg3588 7 күн бұрын
f(x) = (4^(x+5))^(1/(x+3)) has an interesting property: it has both a limit (from below) and a vertical asymptote (from above) at x = -3. I don't think I've seen many functions with this behaviour!
@AFSMG 8 күн бұрын
Realizo todos los ejercicios que propone
@Mephistel 9 күн бұрын
I tried it from the thumbnail after worrying a bit. This wasn't so bad, it's mainly just whether you know the properties of exponents or not. Thanks for the video.
@KasyannoEZMath 9 күн бұрын
You're very welcome.
@michaelcolbourn6719 12 күн бұрын
This is so unnecessary. 3×3×3×3=3⁴=81 So 4=√x and 16=x
@Rando2101 9 күн бұрын
literally the video but without details
@michaelcolbourn6719 9 күн бұрын
Literally not. The video used logs unnecessarily when there is a much simpler way.
@PriscillaMusawa 14 күн бұрын
What if they were mixed fractions
@KasyannoEZMath 14 күн бұрын
Convert the mixed numbers into fractions, then proceed to addition or subtraction.
@PriscillaMusawa 14 күн бұрын
How did you find 5/5 and 3/3
@KasyannoEZMath 14 күн бұрын
I chose 5/5 and 3/3 to get the desired product of 15 in the denominator. Since these two fractions are equal to one, this will not change the value of the number that you multiply with it. In looking for a common denominator, like 15, you can multiply 3 and 5 to make it 15. Once you have a common denominator, then you will just add or subtract the numerator and write down or copy the common denominator.
@jamespn 16 күн бұрын
The answer Is X=0 because the exponents 3^0=1 and 4^0=1, and the bases of 4^1=4, 3^1=3, 4-3=0.
@RedAvalanche98 19 күн бұрын
teaches better than a 30 min lesson
@KasyannoEZMath 19 күн бұрын
Thanks!
@AlexanderSemashkevich 20 күн бұрын
x= -½ 4^(-½)=1/sqrt4=½ ½+(-½)=½-½=0
@b213videoz 22 күн бұрын
this I can't 😂🤪
@KasyannoEZMath 19 күн бұрын
You can't, but you can learn how to solve it.
@b213videoz 22 күн бұрын
4x = x³ 4x - x³ = 0 x(2² - x²) = 0 x1 = 0, proceeding... 2² - x² = 0 = (2-x)(2+x) (2-x) = 0, x2 = 2 (2+x) = 0, x3 = -2 Answer: x = [0, 2, -2]
@KasyannoEZMath 14 күн бұрын
Excellent, Andy O!👍
@michaelcolbourn6719 23 күн бұрын
x³-4x=0 x(x²-4)=0 x(x-2)(x+2)=0 x=0, 2, -2
@murdock5537 23 күн бұрын
Great one, many thanks, Sir!
@KasyannoEZMath 21 күн бұрын
You're very welcome.
@steveeesfan865 24 күн бұрын
2^2.5÷5/4 2^2 4
@KasyannoEZMath 24 күн бұрын
Very good 👍!
@Mark-fl3kx 25 күн бұрын
I must be missing something. Or doing something wrong. because I get approximately 2.973?
@KasyannoEZMath 24 күн бұрын
Go over your solution to find out.
@lemon3015 26 күн бұрын
Great video!!
@KasyannoEZMath 26 күн бұрын
Thanks!
@carloalbertocolaiacovo4182 27 күн бұрын
We can also use lambert w function
@intheinternet8597 Ай бұрын
=+2
@ZainAhmed456 Ай бұрын
X = 1
@LoopRetentionMedia Ай бұрын
Brh 1
@adgf1x Ай бұрын
x=log 5 base3.ans
@KasyannoEZMath 26 күн бұрын
Very good!
@VividM3m0ry Ай бұрын
Difference of cubes: (x-5^(1/3))*(x^2+5^(1/3)*x+5^(2/3))=0
@nasalimbu3078 Ай бұрын
√25 5
@Menacé0nth3g0 Ай бұрын
So goin' to solve the problem before watching the vid, 3^X = 27, by making the base the same, so it has to be 3^X = (3)^3, since 3^3=27, now the base are the same, I can both cancel, the 3, leaving me with just x=3, so now, substitute x=3 and plug in, so 5(3) = 15, so the value of x has to be 15!. EDIT: Got it right!
@KasyannoEZMath Ай бұрын
Excellent solution! Keep up the good work.
@pepebriguglio6125 Ай бұрын
(ln(ln(3))-(ln(ln(4)))/(ln(3)-ln(4))
@matthewfeeley Ай бұрын
Thanks
@KasyannoEZMath Ай бұрын
Thank you so much!
@AreYouDead77YT Ай бұрын
3^x=5 x=log(subscript 3)5 x=log(5)/log(3) x≈1.465
@KasyannoEZMath Ай бұрын
Excellent! Keep up the good work!
@PranjalBose-k6e Ай бұрын
@@KasyannoEZMathwhat is the base of log here e or 10 ?
@MathTheoxist 25 күн бұрын
log(x) means it has the base of 10. Natural logarithms are written as "ln". This is the one in the base of e. @@PranjalBose-k6e
@PranjalBose-k6e 25 күн бұрын
@@MathTheoxist ok thanks
@sajidrafique375 Ай бұрын
you can use any base but log usually is conventionally base 10
@StevenTorrey Ай бұрын
You really have to work out the complete math procedure to get the answer. log 5 (.6989) / log 3 (.477) = 1.4649. And sure enough, proof: 3^1.4649 = 4.9099 = 5.
@xvoidx_yt1723 Ай бұрын
But you’re using a calculator here
@sepehrhaghverdi8977 Ай бұрын
me:0
@도언-x5p Ай бұрын
We can’t solve this equation in real range.
@PawełDąbrowski-n3w Ай бұрын
x = log5/log3 x ≈ 1,465
@Nemazares Ай бұрын
@@PawełDąbrowski-n3w thx
@chandumaddasani9841 Ай бұрын
How can you say a and b are integers
@sebastienlecmpte3419 Ай бұрын
Yeah... Comments before me already pointed out: m=29, n=1 as a solution and I'm adding m=1, n=-27 as well as m=1/27, n=-1
@sebastienlecmpte3419 Ай бұрын
Ok cool. After watching the video I can say... sure if you ever stumble upon a teacher that says "find one of the solution for this equation that should be enough" this way works... m=29, n=1 is also a solution m=1/27, n=-1 is another solution. m=1, n=-27 yet another. And that's before I tried to actually calculate a way to find more solutions.
@Eturoel Ай бұрын
What happened to the Algebraic Rule that to SOLVE an equation every Unknown requires a separate equation; since the given equation has 2 Unknowns, 2 equations are required to SOLVE the original equation to be solved. To find an Answer rather than solve the equation, one could have assumed sets of numbers n,m until the right combination(-s) had been found. Since we are dealing with very small numbers, this would not have taken very long. I am wondering whether you agree with me ? I clearly distinguishing between systematically Solving an equation and simply looking for an ANSWER. u
@LTG_Lanny Ай бұрын
The part inside the parentheses in your final answer can be simplified to log base 4 of 3
@nikotakai8796 Ай бұрын
Why?
@LTG_Lanny Ай бұрын
@ change of base thrm. It’s defined as (ln a)/(ln b) is equal to log base b of a.
@nikotakai8796 29 күн бұрын
@@LTG_Lanny wyld. Thanks.
@AbhishekKumar-uw8hj Ай бұрын
Thnx God! For letting my soul remain for one more day!I mean when people chant"once more" for an artist splendid performance,I say out loud to my life"Heyy!Plz Once more!" On this stage of life where everyone perform it's seven acts! A lot of a big applause for this great life's blood pumping ❤️💜! 💪
@xnicel7364 Ай бұрын
with what base are the logarithms?
@Stevens383 Ай бұрын
Log is always base 10 by default.
@LTG_Lanny Ай бұрын
The common log has a base of 10
@ShootNowAskLater01 Ай бұрын
Common log base is 10. If you’re using a scientific calculator without the ability to alter the base, just do log(x)/log(base) Like to do log base 3 of 9-> (log9)/(log3)
@fullfungo Ай бұрын
Doesn’t matter. The result stays the same
@jajaa8225 Ай бұрын
In higher math its common for the natural logarithm to be denoted as log, but in this case, its probably base 10 logarithm. But it doesn't matter because of the logarithmic rules.
@Sekar-888 Ай бұрын
2^×(1+2^5)=132 2^x(33)=132 2^x=132/33 2^x=2^2 X=2//
@Nemazares Ай бұрын
So 15?
@Xeptor_. Ай бұрын
No
@Nemazares Ай бұрын
Yes, I know but wondering the answer
@goodbye6490 Ай бұрын
No If x is 15 that would mean 3 to the power of 15 equals 5 3 to the power of 15 is 14348907 Which would mean 14348907 = 15 Which it doesn’t
@Jonasz314 Ай бұрын
The math behind this (e.g at timestamp 6.37) is completely wrong. You make the false assumption that (a+b) and (a-b) are integers, when a is a is a square root of an integer, so by definition can be irrational. And of course you can multiple two irrational numbers to get an integer.
@sebastienlecmpte3419 Ай бұрын
Even without irrational I can think of a solution that has non-integer solutions: m=1/27 n=-1
@SantiagoPerezG Ай бұрын
Well even considering only whole numbers, 29^1 - 1^29 = 28 so clearly the solution is incomplete.
@SG49478 Ай бұрын
The reason is that the way he solved it simply does not work to get all solutions and is therefor incorrect. The problem here is that when he made the substitution for a and b, he assumed that a and b necessarily need to be integers. Otherwise the factorization and reducing the possibility to 3 cases does not make sense. However to produce a difference of squares he had to introduce a square root which also can be irrational and therefor not an integer at all. And that's why he lost the solution m=29 and n=1, because a:=29^1/2=sqrt(29) which is irrational making neither a+b nor a-b integers, which is what his way to solve it is built on.
@sebastienlecmpte3419 Ай бұрын
@@SG49478 Without really trying to have a way to calculate all answer I can already think of two more: m=1, n=-27 m=1/27, n=-1
@murdock5537 23 күн бұрын
@@SG49478 Good one, many thanks!
@SG49478 23 күн бұрын
@murdock5537 His way to solve it finds all the solutions for m and n being both even. The case one of them being even and one odd does not work because every power of an odd number is odd again and every power of an even number is even, therefor their difference would be odd in that case and therefor can't be 28. I don't have a good idea, how to find all the solutions for both m and n being odd, 29 and 1 is obvious, I suspect that is the only one, but can't prove it for now.
@murdock5537 23 күн бұрын
@@SG49478 Agreed. n^2 = 8√5; m = 3 → m^n ≈ n^m + 28
@JohnSmith-nx7zj Ай бұрын
You didn’t specify at the start that we were looking for integer solutions…
@sebastienlecmpte3419 Ай бұрын
Yeah I when I see these I like to try and solve before watching the video so any restriction explained only after the video starts really bugs me. Quickly without doing any real math, I see two answers that jump out, although one does not respect the "integer solution only" rule: 29¹-1²⁹=28 (1/27)^(-1)-(-1)^(1/27)=28 I'll try to solve in a more general way and then watch the video.
@JohnSmith-nx7zj Ай бұрын
@ how did you get on? You found two solutions that the video didn’t, and furthermore one of your solutions uses just integers. My instinct on seeing this problem is that it’s probably an exceptionally difficult problem to solve if you assume integer solutions. It’s essentially a difference of perfect powers so I’d assume along the lines of Catalan’s conjecture, albeit a special case. If you’re just looking for real solutions it seems fairly obvious there’ll be infinitely many, which just involves expressing m in terms of n or vice versa.
@b213videoz Ай бұрын
This one is scary
@KasyannoEZMath Ай бұрын
Really?
@b213videoz Ай бұрын
@KasyannoEZMath yes! nested log()
@Menacé0nth3g0 Ай бұрын
ln5/2? Is my answer. (I texted this, before watching the vid.)
@KasyannoEZMath Ай бұрын
Very good!
@1234larry1 Ай бұрын
The equations for the check are (3+sqrt5))/2)+((4+(2sqrt5/)2)=7+((3sqrt5))/2