Video

Not a number

100

❤❤❤

-80

So obvious when tutor explains. I could see that 729 = 3⁶ but forgot the obvious tactic of getting the 2 bases equal: 9 ^ ( 2√x ) = 9³. Practicing algebra requires great peace of mind. 😊 Excelsior!

I can confidently say that 90percent did not fail. (People do this in 3rd grade).And also its -13

_17

-13

❤❤❤❤

Error, 2(5-5)=3(5-5), it's 0=0

Hmm... If we anticipate - as we do at (2:50) - that the intermediate solution should consist of integer numbers, we could spare all the steps in between and directly go from the term written at (2:55) to the solution. The solution even directly jumps into our eyes since we have the "5" and the "2" already in the original terms. When we then set sqrt(5)^x == 5^2 and sqrt(2)^x == 2^2, it so happens that x comes out as 4. If we would NOT anticipate the intermediate integer solution beforehand, but instead solve for a general case of arbitrary bases on the LHS and an arbitrary value on the RHS, we would have to fall back to some approximation techniques. Which would as well not be that much of a burden, since we can guess a good starting point by just probing some integer values for x. In fact, doing so results in the shortest possible road to Rome in this specific case - without all the hassle of refactoring and intermediate solution at all -, since we pretty directly land at the correct solution immediately (because it is, in fact, an integer). Since the sqrt(5) part is the biggest one, we can estimate the coarse range of the contribution of that term by directly estimating the log(something around 20) / log(5), which yields 2, which in turn yields 4 for x. Entering that shows it to be the correct solution already. What would be interesting: Are there different roads to the solution allowed by the presentation of the exercise at the olympiad? Or do they always require the same (lame, because of using anticipation anyways) way?

Mashallah ❤❤❤❤❤

❤❤❤❤❤❤❤

Math Olympiad: 144^√x = x^12; x = ? (x^12)^[1/(12√x)] = (144^√x)^[1/(12√x)] x^(1/√x) = 144^(1/12) = 144^(1/√144); x = 144

saved it for later

Aapko Dil Sy salam sir G

Sorry. 10(0) = 0. Nice trick though.😅

🙍‍♂️+🙍‍♀️=👶 3 total

3^2×2-2^2×1 x=2,y=1

Lgta he English mein fail ho, abbreviation hota he...

Hi

Why so complicated? Just gather at one side of the equation, getting (5/7)^x == 1, yielding x == 0 at first glance (or just glance it without any reformation - it's really trivial). No exponential curve goes through the same value of y twice - except if the base is exactly 1.

10(11-11)=11(11-11) (10×0)=(11×0) Increase your iq level

"Promosm" 😉

sir ji Mai Aapki har video ko like krti hon aur Aapki video ka intazar rahta

What you do from 3:30 onwards, is a stipulation that all the roots are just distraction flares and the solution you go for shall consist only of integers. Which admittedly is not a wrong idea, since roots are generally irrational except the few cases were they are not. BUT: If you stipulate an integer solution ANYWAYS, you could have got on that road MUCH earlier, right from the beginning, by substitution. All that rewrite and squaring again was a complete waste.

Just czcams.com/video/pt7sktAGlzg/video.html

nice for a morning exercise but works as well for x= -9 and y= -8 then for (-9;8) and (9;-8) because of |x| and |y| lol prof ! ayala & yami, shamanes associate

Use fixed-point iteration, x←1/5^x, starting with x=1/2 to obtain x=0.469621922936... This form of FPI equation converges slowly. An alternate form may converge faster or may blow up. If x←f(x) represents an FPI equation, then |f'(x)|<1 determines convergence rate where f'(x)=∂f(x)/∂x and |•| represents absolute value. The smaller |f'(x)| is the faster the convergence rate.

Use fixed-point iteration, x←2^(1/x), starting with x=1.5, to obtain x=1.559610469462369... on my pocket calculator. How was x=e^[W(ln2)] evaluated? You just wrote down 1.5596 and didn't explain where it came from. That's no way to teach. Did you use Wolfram-Alpha to evaluate W(ln2)? If that is the case then why not just enter solve x^x=2 and it returns x=e^W(ln2)=1.55961? What did I learn from that? Nothing. To solve real-world problems, fixed-point and Newton-Raphson iteration are the best choice.

100 caror

x^ln(x)=x^5 ln(x)ln(x)=5ln(x) ln(x)=5 x=e^5 x=148.413159103 Check: (148.413159103)^ln( 148.413159103 )=(148.413159103)^5 72004899339.4=72004899338.4

Solving problems like this is often easier using a simple trick. Memorize it. r=floor(ln42/ln2)=5. q=floor[ln(42-2^5)]=3. p=ln(42-2^5-2^3)/ln2=1. Try solving this problem: 7^p-5^q-3^r=820391, p>q>r, which seems hopeless. In this case we see that 7^p must be more than 820391, so use the ceiling function: p=ceiling(ln820391/ln7)=7. Then 5^q+3^r=7^p-820391, and 5^q must be less than 7^p-82039, so use the floor function: q=floor[ln(7^7-820391)/ln5]=5. Finally r=ln(7^7-820391-5^5)/ln3=3. This method works only if the exponent of the largest remaining term is solved for at each next step, which is the way this type of problem is usually set up. In the given problem, we know that 2^r>2^q>2^p, so solve for r first, q second, and p third. In my example, we know that 7^p>5^q>3^r, so solve for p first, q second, and r third.

5^x=1/x 5^x•x=1 x•e^xln(5)=1 xln(5)•e^xln(5)=1•ln(5) W(xln(5)•e^xln(5))=W(ln(5)) xln(5)=W(ln(5)) x=W(ln(5))/ln(5) x=0.4696219229356105441178030668907995994378529299815519288056339238 Check: 5^( 0.4696219229356105441178030668907995994378529299815519288056339238)=1/0.4696219229356105441178030668907995994378529299815519288056339238 2.12937248276=2.12937248276

Hmmm... This coarse result is around 5% off, thus really only a coarse estimate.

This method ignores other real solutions if they exist. In this problem, 1.2396277... is also a solution. It requires the Lambert W function to find this (or a numerical search).

Equate powers of 3: x-y=2 and x+y=3; 2x=5 yields x=5/2 and y=x-2=1/2

9(11/10)^2 vs 8(11/10)^3 or 9 vs 8(11/10) or 9 vs 8.8, so (3.3)^2 is larger.

Substitute y=2^k into the given equation and rearrange to y^2=16, giving y=±4. For real k, reject y=-4, so k=ln4/ln2=2.

Split this problem for easier solution: x=z+5/2 and y=z-5/2; (z+5/2)^2+(z-5/2)^2=2z^2+25/2=17 z^2=(34-25)/4=9/4; z=±3/2; x=z+5/2=±3/2+5/2=4 or 1 and y=z-5/2=±3/2-5/2=-1 or -4.

ln[x^(1/2)]=ln(x)/2=√[ln(x)]; ln(x)^2=4ln(x); ln(x)=4; x=e^ln(x)=e^4=54.598...

❤❤❤❤❤❤❤❤

This is only half the way to the solution, since evaluating the Lambert function is not at all "business as usual". This function is up to day NOT INCLUDED in most pocket calculators, in most math packages for programming languages, or in table calculation programs like OpenOffice Calc. Thus, to evaluate it, you would have to resort to Wolfram Alpha or Mathlab or the like services. IF you are forced to use THOSE, you can put in the original expression straightaway. Thus, at least at the moment, the use of the Lambert function is still somewhat limited and exercises around its use are somewhat lame. You have at least to refer to the math library to use for resolving the function. (Python, for example, has one, PHP not, Javascript also not, for C++ and Java you can get one, but not in the standard libraries and so on...) Where you don't have that function build in, you have to manually program the approximation by something like newtons method and you have to solve all the details around that, too, as long as you want something remotely generally reusable. All that on top of that you have to be firm with programing and math at an engineering level and you have to dig in the appropriate public documentation for the approximation of the function... Pretty bloated stuff before you are able to solve that, isn't it? And lastly: If we are forced to approximations anyways, and forced to manually build them anyways, then we could equally well put the original expression without any further reformations (except giving the first derivative) directly into a general implementation of the newton method. That gives a drastically shorter way to Rome! :D

Mashallah❤❤❤❤❤❤❤❤❤

In this case, i disagree: If you handle the root of "i" as possible "+45°" or "-135°" rotated, then you have to handle BOTH roots independently. Thus you may have for sqrt(i): unit vector "+45°" or unit vector "-135°", and you may have for sqrt(-i): unit vector "-45°" or unit vector "+135°". In this case, the solutions "+/- sqrt(2) * i" enter the stage, since you may combine the unit vectors of both roots in 4 different ways. In practical applications, you may of course have conditions where the allowed rotation of the roots is constrained. But then even the solution "-sqrt(2)" may become forbidden.

Why such a complicated approach? Just taking log on both sides in the original equation yields the exact same result. To split the 33 into 11 * 3 in between is an additional redundancy on top of the complicated deduction road. You do not save up operation number or operation costs. Just log(33) / log(3) - and you are done. At the same time: Nice to see all roads ending up at the same spot in Rome :D

This incremental approach is the way you can tackle any such problems with real numbers as well. As soon as you realize that an integer solution is not trivially in reach, you can switch to the generalized version (newtons method) with the best so far integer approximation as the starting point.

This equation has five roots, x=0, x=0, x=1, x=(-1+i√3)/2 and x=(-1-i√3)/2 How is that? Rearrange the equation to x^2(x^3-1)=0. Either x^2=0 or x^3-1=0 The latter equation may be expressed as x^3=1=e^(i2jπ),j=0,1,2, and its three third roots expressed as x_j=[e^(i2jπ)]^(1/3)=e^(i2jπ/3)=cos(2jπ/3)+i*sin(2jπ/3),j=0,1,2

6=3

4