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")

@jensraab2902

15 күн бұрын

Yes, can't always rely on luck. Also, you didn't rule out answer d, three solutions, right?

@deltalima6703

15 күн бұрын

Nope might be d. Doubt it. Maybe 10% chance.

@freaze2048

14 күн бұрын

@@jensraab2902 yea i saw answer D but I was just like "it's 50/50 so ill just go with C"

@jensraab2902

14 күн бұрын

@@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.

@freaze2048

14 күн бұрын

@@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 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 bet that is a "Make sure you read the question properly" question.

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

@DergaZuul

12 күн бұрын

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.

I have a doubt problem......integral of ((1-x^7)^1/4 - (1-x^4)^1/7) can you please solve this BPRP?

@Misteribel

15 күн бұрын

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?

@Patrik6920

15 күн бұрын

Note .. between 0 and 1 the integral must be zero (the only real solution)

@sinekavi

15 күн бұрын

@@Misteribel Let me rephrase the question......i have a doubt in that problem

@sinekavi

15 күн бұрын

Edit: I want BPRP to solve this problem

@Patrik6920

14 күн бұрын

@@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...

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).

You can even use logarithm of base 2 to find the answer.

Sir Please explain limitation of descartes rule of sign

@ValentineRogue

15 күн бұрын

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

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.

I just made all the terms base equal to 2 and then compared the powers

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?

@Apollorion

46 минут бұрын

excuse my too quick responses.. 8^x + 4 = 4^x + 2^(x+2) = 2^(3x) + 2^2 = 2^2x + 2^(x + 2)

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

@Nigelfarij

15 күн бұрын

Why must x be an integer.

Identity of 'x' ?

Teacher plz teacher Integrity [sinx+sinx tanx + secx]/[sin^2x tan^2x + 2(1+tanx)] dx

@anigami01

16 күн бұрын

let me do it it seems to be a good exercise for me

@tglmkyoutube

15 күн бұрын

@@anigami01 so bro have you finished?

👍

asnwer=(b) one

B and C are both correct. The equation has one real solution. It also has another real solution.

i put x=0 and x=1 and they satisfied equation and .......

@55hzdxlh73

15 күн бұрын

imagine the eqt has 3 roots😂

@matta5749

15 күн бұрын

that doesnt answer the question. you still need to know how to prove that there are no other solutions, otherwise you're just guessing.

@geirmyrvagnes8718

15 күн бұрын

@@matta5749 Eliminating two options on a multiple choice test is helpful, I guess.

@jensraab2902

15 күн бұрын

@@geirmyrvagnes8718 Is it really? This is Oxford. They want to see that you know what you're doing, not just doing lucky guesswork. 😉

@geirmyrvagnes8718

15 күн бұрын

@@jensraab2902 Getting things roughly right in a fraction of the time? Yeah, maybe not if you are supposed to study mathematics. 😅

First

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