A Homemade Exponential Equation

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[ilana]

Dear sir! You are an excellent teacher. Great examples but a bit too fast. Please some examples for less advanced people.. Thanks in advance!

[tambuwalmathsclass]

Gather here if you have realized 2 as a solution before solving 😁

[armacham]

Another way to solve it is to recognize that if you make a new variable a = x/2 then you can restate the equation as:

5^(2a) + 5^(2/a) = 50

which is the same as

25^a + 25^(1/a) = 50

from there it is obvious that a = 1 is a solution and it's not too difficult to prove that there are no other solutions.

[prakrit]

Genious Sir👍😎👍Gained a new way of approaching maths. Thank you for the informative videos, Sor

[copernic]

The other method is:
1) subtract 50 to both sides
2) multiply by 5^(x/4)
3) t = 5^(x-2)
etc.

[scarletevans]

One of my favourite inequalities, also based on AM-GM, is that if ab>0, then
a/b + b/a >= 2 (and its generalization for n (see below))
Thus, 2/x+x/2 >= 2
4/x+x >= 4
This can be generalized in a cyclic way for x1, ..., xn, i.e.
x1/x2 + x2/x3 +...+ xn/x1 >= n
Proof is super easy with AM-GM :D

[vaggelissmyrniotis]

You can take f(x)=5^x +5^(4/x) and find with the first and second derivatives that is decreasing for 0<x<2 and increasing for x>2 so the minimum value of f is at x=2 f(2)=50 which is the only solution.for x<0 we have f(x)<2 so there is no solution if x<0.

[eckhardfriauf]

.. so nice to see a video in which the geometric means plays a role! Thanks.

[goldfing]

As usual, we have s simple integer solution, x = 2, because
5^2 + 5^(4/2) = 5^2 + 5^2 = 25 + 25 = 50.

[rakenzarnsworld]

Hint: 25+25=50
5^x = 25
5^(4/x) = 25
x = 2

[dwarfy]

Just multiplying both sides by 5^x, substitute 5^x by X I get the second order polynomial X²-50X+(25)²=0 (5^4=5^2^2), perfect square of (X-25)². Therefore 5^x=25, unique solution : x=2
Edit : error, do not take this answer into account

[forsee]

x approaches 1 on negative inf does not prove y can't reach 50 on x<0 does it?

[flarklooney]

UGH! My eyes! Please Sir, can we have the darkmode thumbnail back? 🙏
Thank you for uploading these videos.

[oki]

i looked at the thumbnail and immediately knew x = 2, can someone tell me what all of this is for though?

[walterwen]

A Homemade Exponential Equation: 5^x + 5^(4/x) = 50; x = ?
5^x + 5^(4/x) = 50; 50 > 5^x, 3 > x > 0
1. First method:
5^x + 5^(4/x) = 50 = 25 + 25 = 5^2 + 5^2; x = 2 and 4/x = 2, x = 2
2. Second method:
[5^x + 5^(4/x)]/25 = 5^(x – 2) + 5^(4/x – 2) = 50/25 = 2 = 1 + 1 = 5^0 + 5^0
x – 2 = 0; x = 2 and 4/x – 2 = 0; x = 2
3. Third method:
Let: y = 5^(4/x), y^x = 5^4
5^x = [5^(4/x)]^[(x^2)/4] = [y^(x^2)]^(1/4) = [(5^4)^2]^(1/4) = 5^2; x= 2
5^x + 5^(4/x) = 5^2 + y = 50, y = 5^(4/x) = 50 – 5^2 = 5^2, 4/x = 2; x = 2
4. Fourth method:
5^x + 5^(4/x) = [5^(4/x)]^[(x^2)/4] + 5^(4/x)
= [5^(4/x)]{[5^(4/x)]^[(x^2)/4 – 1] + 1} = 50 = (5^2)(2)
If: 5^(4/x) = 5^2, 4/x = 2; x = 2 then [5^(4/x)]^[(x^2)/4 – 1] + 1 = 2
[5^(4/x)]^[(x^2)/4 – 1] = 1 = [5^(4/x)]^0, (x^2)/4 – 1 = 0, (x^2)/4 = 1, x^2 = 4; x = 2
Final answer:
x = 2

[moeberry]

This was an easy problem however you have to make it more difficult Syber by not putting the minimum as y=50 your just calculating the minimum value of the function and setting it equal to the value of the function at that same x value. If the horizontal line was above 50 you would have 2 solutions and if below but greater than 1. You would have no solutions. AM-GM limits your domain and that’s why you don’t get the second piece of the graph unless you look at limits to determine vertical asymptotes and horizontal asymptotes. These types of problems are made by first finding the derivative and finding it’s critical points in this case they are minimum values and you then set the function equal to that minimum value. In this case you set it equal to 50 but you could have also set it equal to 2/25 which is the second minimum value and would still give only 1 solution.

[tetsuyaikeda]

you say that I have to use arithmetric and geometric mean relation ??