Solving an exponential equation from Oxford
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- Опубликовано: 15 май 2024
- We would like to know the number of real solutions to the exponential equation 8^x+4=4^x+2^(x+2). This question is from the University of Oxford Math Admission Test in 2007. www.maths.ox.ac.uk/system/fil...
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I just looked at the thumbnail and randomly tried to replace x with 0 and with 1 and luckily solving this in like 1 minute (even tho it's not really a good way of justifying saying "it's evident")
Yes, can't always rely on luck. Also, you didn't rule out answer d, three solutions, right?
Nope might be d. Doubt it. Maybe 10% chance.
@@jensraab2902 yea i saw answer D but I was just like "it's 50/50 so ill just go with C"
@@freaze2048 Fair enough. But this is why I personally think multiple choice questions are a total nonsense in math. I don't know if this is an American thing; I don't think I've ever seen this in a math test in my country.
@@jensraab2902 For my knowledge I've also never seen this type of math test in France
0:15
"Try this first."
*"No."*
Sir , I am you're big fan from India 🇮🇳 and I love ❤ you're basic mathematics videos
Sir , can you plss solve this exponential equation ,
6^(x) - 1 = 4^(x) + 2^(x/2) - 3^(2x)
Find the " x " and number of solutions for this equation.
Sir , plsss make a full video on solving this exponential equation
Lot of love 😘 & support 🫂 from me !!!!!!!!!!
I rearranged terms and factored it.
4^x (2^x -1)=4(2^x -1)
Then either 4^x=4, so x=1, or 2^x -1=0, so x=0.
I bet that is a "Make sure you read the question properly" question.
i have turn the equation to be 2^a + 2^b = 2^c + 2^d
and from that we can have a set of logic statements:
a = c and b = d
or
b = c and a = d
wich gives the same solution
I managed to factor it to (4^x - 4)(2^x - 1), giving the two real solutions x=0,1. I realise now that 4^x-4 is a difference of two swuares and can be further factored, but that leads to the non-real solition
Well to simplify those are monotonous growing function minus constant in both terms. At most can have one real solution so easily see that one in each term so no need to think of difference of squares. I did almost same way just guessed 0 and 1 in first 15 sec and then justify that no other real solution is possible.
You can even use logarithm of base 2 to find the answer.
Another way to explain why 2^x = -2 is not possible as a real number is because the range of the exponential function is (0, infinity).
Sir Please explain limitation of descartes rule of sign
Descartes' rules of signs is that the number of real solutions to a polynomial is equal to the number of sign changes of the polynomial when written in descending degree order or less than that by a multiple of 2. When bprp writes the equation as "t^3-t^2-4t+4=0", there are 2 sign changes, so there can only be 2 or 0 real solutions
another possibility is to realize that t=2^x must be a power of 2, and quickly realise t = 2 is one of the solutions of 2. then do polynomial devision of (x - 2), and find a quadratic that we can solve from that
Why must x be an integer.
👍
I have a doubt problem......integral of ((1-x^7)^1/4 - (1-x^4)^1/7) can you please solve this BPRP?
What is a doubt problem? Do you mean you doubt whether there's a problem, or you doubt the solution (which you didn't give yet)? Try asking on Math SE perhaps?
Note .. between 0 and 1 the integral must be zero (the only real solution)
@@Misteribel Let me rephrase the question......i have a doubt in that problem
Edit: I want BPRP to solve this problem
@@sinekavi probably be a very long video .. maby he has somewhere u can reach him .. unless he has a cheat up his sleeve .. the solution will be many many pages long...
I tried solving it by rewriting in terms of powers of 2: 8^x + 4 = 4^x + 2^(x+2)
2^(3x) + 2^2 = 2^2x + 2^(x + 2). When you solve this, you find that every value of X satisfies the equation. What did I do wrong?
excuse my too quick responses..
8^x + 4 = 4^x + 2^(x+2) = 2^(3x) + 2^2 = 2^2x + 2^(x + 2)
Teacher plz teacher
Integrity [sinx+sinx tanx + secx]/[sin^2x tan^2x + 2(1+tanx)] dx
let me do it
it seems to be a good exercise for me
@@anigami01 so bro have you finished?
I think t substitution is not necessary. You can factor 2^2x from 2^3x and 2^2x and factor -4 from the rest. You get 2^2x(2^x-1) and -4(2^x-1). Then factor (2^x-1) to get equation (2^2x-4)(2^x-1)=0.
Identity of 'x' ?
B and C are both correct. The equation has one real solution. It also has another real solution.
Intuitively before watching: This is a cubic in terms of 2^x, so expected to be 3 real solutions, by observation 1 and 0 are solutions, my guess is the third would be 2^x = negative which isn't real, so answer is c
i put x=0 and x=1 and they satisfied equation and .......
imagine the eqt has 3 roots😂
that doesnt answer the question. you still need to know how to prove that there are no other solutions, otherwise you're just guessing.
@@matta5749 Eliminating two options on a multiple choice test is helpful, I guess.
@@geirmyrvagnes8718 Is it really? This is Oxford. They want to see that you know what you're doing, not just doing lucky guesswork. 😉
@@jensraab2902 Getting things roughly right in a fraction of the time? Yeah, maybe not if you are supposed to study mathematics. 😅
First