How many real solutions does this logarithmic equation have? (Oxford MAT)

Рет қаралды 7,350

Күн бұрын

Пікірлер: 30

@cyrusyeung8096 4 ай бұрын

For the no real solution suggestion, you can just add negative sign inside both log, i.e. log2(- 2x³ - 7x² - 2x - 3) = 3log2(- x - 1) + 1

@xinpingdonohoe3978 4 ай бұрын

I guess so. On the left you'd have +πki/ln(2) and on the right +3πki/ln(2) for some odd integer k (therefore non-zero), so it wouldn't be really feasible to try and equalise them.

@JonnyBoi957 3 ай бұрын

I haven't even done log, and I aparantly didn't even need a log to solve this. It is way easier than it seems. When I saw you making both log 2, I thought yeah, the brackets are going to be the same. Thank you for explaining so well.

@piccolo64 4 ай бұрын

i did the same🙂 nothing like solving a math problem to feel happy and satisfied🥰

@noneelno 4 ай бұрын

Just a suggestion, I would love you try and solve a HSC Maths extention 2 paper in similar to the GSCE A level video you did previously.

@PauloChacal 3 ай бұрын

Agree one root could be negative and demand interpretation. It would be less obvious.

@tobybartels8426 3 ай бұрын

Even if there was some negative coefficients in that cubic polynomial, you could still trust that both solutions work in the real-number system as long as both values of x+1 are positive (because then 2(x+1)³ is also positive and that is equal to the complicated cubic polynomial for these values of x).

@riccardovianello7710 3 ай бұрын

I'm totally disturbed by the way you solved that quadratic equation. I'm so used to X1,X2= (-b+-sqrt(b^2-4ac))/2a.

@brandonramnarine2410 4 ай бұрын

you should check out CAPE pure mathematics unit 1 or unit 2

@saviplayer4546 3 ай бұрын

Aye I believe ik you

@psychoranzer2495 3 ай бұрын

We still have to check the domain first to verify

@EpikXeuxy 4 ай бұрын

great video! pls solve some maths jee questions.

@K2MusicKSquare 4 ай бұрын

Wouldn't the negative answer still be a valid solution if the both sides get a complex result that are equal, and the negative answer is still real?

@hafizusamabhutta 4 ай бұрын

Masha sense.

@oryxisatthefront8338 4 ай бұрын

How many REAL solutions x are there to the following equation?

@xinpingdonohoe3978 4 ай бұрын

@@oryxisatthefront8338 yes. We want real x. But that doesn't impose on the original. For example, how many real solutions to √x=i are there? We find real x that satisfies. We aren't asked to find real x that satisfies and also keeps the original expression in R.

@Brid727 4 ай бұрын

the question never asked you to find all values of x, it only asked to find the number of REAL solutions that satisfy the equation once you find that, going beyond that is just a waste of your time so yeah but of course it would be a valid step if it were that the question asked to find all values of x

@xinpingdonohoe3978 4 ай бұрын

Yes. If they are real and solve it, that's good. Even if proving that they solve it means delving into the complex plane, they still solve it.

@Verifyourage 4 ай бұрын

Sometimes, very elementary maths is required . Just draw an x,y table and join the dots. And presto There's your cubic😂😂😂😂😂

@ronbannon 4 ай бұрын

Rewrite: Find the values of $a$ such that the following equation has two real solutions and only one real solution. $\log_2 \left( 2x^3+7x^2+2x+a ight) = 3 \log_2 \left( x + 1 ight) + 1$

@robertveith6383 3 ай бұрын

Don't bother to write Latex notation. It won't show up in special text in a KZbin post, and, consequently, it's harder to reader because it is more messy.

@creamyscroll2485 3 ай бұрын

this problem made me feel like I am good at math and I also can give this admission test. psss, nah! I suck at math, it was just the fact that this easy problem gave me overconfidence. Sorry, lol🤣🤣

@simoneantoniocarretta1048 4 ай бұрын

if: ... =3 log(x - 1) + 1... NO real solution... 🤔

@David-cd7ip 4 ай бұрын

I suppose that even if the solutions are negative, they still are real solutions. It requires the analytic continuation of logs to make sense of it, but the solutions would technically be real.

@stevemonkey6666 4 ай бұрын

How can you be sure that the +1 at the end is base 2 and not base 10?

@bprpmathbasics 4 ай бұрын

Bc it says log_2

@joaomane4831 4 ай бұрын

Bro... Really?

@firstnamelastname4582 4 ай бұрын

You can basically do whatever you want with that last 1. 1 = log_2(2) = log_e(e) = log_10(10). But here the useful one is 1 = log_2(2) because the other log has base 2

@ronaldking1054 4 ай бұрын

It's arbitrary what 1 equals. There are many expressions. He just chose the one that helped him the best that did not violate any rules of equations. Things that would violate would be log with a base of a negative number of that negative number or square root of -1 times -1. Those are not in the domains of the functions that he is using.