A Nice Exponential Equation | Math Olympiads

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(x^2-7x+11)^(x+1)=1
#ChallengingMathProblems #ExponentialEquations #Exponentials
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

SIMILAR PROBLEMS 🤩

PLAYLISTS 🎵 :

[goldfing]

The left side is certainly = 1 if the base is = 1. So we have to solve
x^2 - 7x + 11 = 1
x^2 - 7x + 10 = 0
(x - 2)(x - 5) = 0
x = 2, thus (1)^(2+1)= 1^3 = 1, true
x = 5, thus (1)^(5+1) = 1^6 = 1, true
Another possibility is that the base is -1 and the exponent is even:
x^2 - 7x + 11 = -1
x^2 - 7x + 12 = 0
(x - 3)(x - 4) = 0
x = 3, thus (-1)^(3 + 1) = (-1)^4 = 1, true
x = 4, thus (-1)^(4 + 1) = (-1)^5 = -1, false
And the base could be <> 0, the exponent = 0:
x + 1 = 0
x = -1
thus
x^2 - 7x + 11
= (-1)^2 - 7*(-1) + 11
= 1 + 7 + 11
= 19
And 19^(-1+1) = 19^0 = 1, true.
So the solution set is {-1, 2, 3, 5}.

[alextang]

This kind of question no longer is a problem in this channel. 😋😋😋

[mustangjoe]

Interesting problem.
I took natural log on both sides leaving me with:
(x+1)ln|x²-7x+11|=0 with the Interesting part being if x²-7x+11 is -1 then x must be in form 2k+1, k€ R
So, x+1=0 --> x=-1
ln|x²-7x+11|=0 gives x=2, x=5 when positive and x=3, x=4 when negative. The form 2k+1 is an odd number so only 3 is valid.
x= -1, 2, 3, 5

[raymondarata]

I used Desmos and it showed a scattering of points between approximately x = 2.4 and 4.6 which happens to be the two roots of the polynomial. One of those scattered points is (3, 1) and it displays that point as a solution to the intersection of the LHS and RHS. along with the other three points. Desmos worked for me.

[drdiegocolombo]

Grande Valerio!!!!
Vogliamo altri video sulla MQ😉

[ManjulaMathew-wbzn]

I figured out if complex solutions are acceptable for x.
Yes!! Take ln of the original equation
ln(x^2-7x+11)^(x+1) = ln1
(x+1)ln(x^2-7x+11)=0
Either x+1=0 or ln( x^2-7x+11)=0
The second condition doesn’t stipulate x has to be real. As long as the complex solutions make the quadratic expression to be equal to 1 the above equation is satisfied. Wen you re substitute the complex solution z in original equation
(1)^(z+1) = (cos(0)+isin(0))^(z+1)
= (cos(0*(z+1))+ isin (0*(z+1))
=cos0 +isin0 = 1
So complex solutions are ok.

[vladimirrodriguez]

Complete analysis of all possible real solutions. Perfect Syber

[miro.s]

And now how to find complex solutions that will probably contain transcendent numbers?

[ManjulaMathew-wbzn]

I figured out what happened to x=3 on the graph. The function f(x) is real only when the base quadratic is positive or the exponent is an integer. If you draw the graph of x^2-7x+11=0 you will see the roots are exactly at the point where f(x) disappears. However at x=3 y=1 and x=4 y=-1 are the only two points f(x) is defined in this interval. They are just two dots on the graph and that’s why not visible. For example when x=3.5 the base is negative and the power involves squarerooting and f(3.5) would be like re^3PIi/2 which is purely imaginary and not shown on a real graph. Interesting!!!

[samf]

If my process is wrong please tell me
Let, X²-7x+11= K
Then K^(x+1)=1
Therefore K^x =K^-1
By using log we can get x=-1

[mcwulf]

Ok, the bracketed number must be 1 or, if x odd, then -1.

If it's 1 then LHS factors to x=5 or x=2.

If it's -1 then x=4 or x=3. Only the second will give 1 when raised to the power x+1.

So X € { 2, 3, 5 }

[stianuwolyre]

I tried finding the imaginary solutions, but unfortunately couldn't figure them out and went on to wolfram alpha. Apparently the imaginary solutions are approximately where x≈-1.41252+1.82344i and x≈-1.41252 - 1.82344i. Wolfram alpha couldn't help me in finding the non approximate forms, and my attempt at putting it in a trigonometric form, imaginary powers, and logarithm stuff seemed out of my level of math. Anyone have any ideas how these solutions are formed?

[zoheirzerguine]

Where is the three, you have approuved that 3 is a solution, but it does not appear on the graph ???

[ianfowler]

Serhii has nailed it. Exponential functions are only defined when the base is positive. Otherwise you get into a quagmire of discontinuities. For example : y = x^x, (-1, -1) satisfies it, but why is it not on the graph?? Because as soon as you take any neighbourhood around -1 all hell breaks loose. Discontinuities abound. The graph becomes completely crazy when x<=0.

Same thing applies here. If you take a close look at the graph in the video you can plainly see that there is an interval along the x-axis where the graph does NOT exist - i.e. these are the x-values where the base is NEGATIVE. 3 just happens to be inside this interval.

Two minute penalty SM - you should have known that. (LOL) .

[quantumbuddha]

Fun problem! I haven't watched the video yet. I see four real solutions: -1, 2, 3, 5.

[forcelifeforce]

WolframAlpha also leaves out x = 3 as a solution.

[renyxadarox]

Exponentiation is not defined for negative bases in the domain of real numbers. And yes, there are no exceptions for integer negative numbers. Case 2 could be valid in the domain of integer numbers, but it was claimed at the very beginning, that we are searching for REAL values of x. Real values of x imply real values of polynome of x, even if they look like integers.

[scottleung]

3 is not in the graph because then the base would be negative, which Desmos can't really articulate.

[mehulpunia]

I have a fun logical puzzle, let's see who can solve this.
There is a conversation between two friends namely Mehul and Rahul.
Mehul: how many children do you have?
Rahul: I have three children
Mehul: what are their ages?
Rahul: the product of the ages of my children is 36
(Mehul did some calculation)
Mehul: I could not figure out the ages.
Rahul: the sum of the ages of my children is equal to the number of buildings around us.
( Mehul counted the buildings but still he don't know the ages)
Rahul again said: my smallest child has Blue Eyes.
Mehul exclaimed because he now know what the ages of the children are.
so can you solve for the ages of the children?

[angelishify]

Izgleda, da je ta kanal promocija za Desmos. Nimam nič proti. Primeri so zanimivi in poučni.