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

Подписаться 134 тыс.

Просмотров 7 тыс.

50% 1

How many real solutions does this logarithmic equation log_2(2x^3+7x^2+2x+3)=3log_2(x+1)+1 have? This question is from the University of Oxford Maths Admission Test in 2022. Check out more videos on the Oxford MAT: • Oxford Math Admission ...
Shop my math t-shirts & hoodies on Amazon: 👉 amzn.to/3qBeuw6
-----------------------------
I help students master the basics of math. You can show your support and help me create even better content by becoming a patron on Patreon 👉 / blackpenredpen . Every bit of support means the world to me and motivates me to keep bringing you the best math lessons! Thank you!
-----------------------------
#math #algebra #mathbasics

Опубликовано:

21 май 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 31

@piccolo64 28 дней назад

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

@noneelno 29 дней назад

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.

@cyrusyeung8096 28 дней назад

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 28 дней назад

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 23 дня назад

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.

@tobybartels8426 25 дней назад

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).

@EpikXeuxy 28 дней назад

great video! pls solve some maths jee questions.

@PauloChacal 19 дней назад

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

@riccardovianello7710 17 дней назад

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.

@psychoranzer2495 24 дня назад

We still have to check the domain first to verify

@brandonramnarine2410 28 дней назад

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

@saviplayer4546 5 дней назад

Aye I believe ik you

@Verifyourage 28 дней назад

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

@K2MusicKSquare 29 дней назад

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?

@hafizusamawrites 29 дней назад

Masha sense.

@oryxisatthefront8338 28 дней назад

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

@xinpingdonohoe3978 28 дней назад

@@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 28 дней назад

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 28 дней назад

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.

@comdo777 28 дней назад

asnwer=1x isit hmm dhmm

@simoneantoniocarretta1048 28 дней назад

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

@stevemonkey6666 29 дней назад

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

@bprpmathbasics 29 дней назад

Bc it says log_2

@joaomane4831 28 дней назад

Bro... Really?

@firstnamelastname4582 28 дней назад

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 28 дней назад

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.

@creamyscroll2485 20 дней назад

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🤣🤣

@ronbannon 28 дней назад

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 23 дня назад

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

@David-cd7ip 28 дней назад

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.