Try an A-Level further inequality: kzread.info/dash/bejne/Y6SMx7aOj8aeZaw.html

After obtaining the equation log base 2 (x^3/2) = 3, you could also put the 3/2 in front of the log expression via property of logarithms. Multiply 2/3 on both sides would give you log base 2 (x) = 3*2/3 = 2. Raise both sides by the base 2, you'll get x = 2^2 = 4 which is pretty much the same answer. Love your videos BTW, more power to you! 祝你一切順利！

@JayTemple

Ай бұрын

That's what I did too.

@Professor_Sargeant_JAMS

24 күн бұрын

Beat me to it.

This is so exciting! Im giving my A-levels in the summer this year (not so far away from now). Suddenly seeing content like this on your channel has been a treat ❤🎉

Got my A Level in June, glad he’s doing A Level questions now 🎉

These are so useful, please keep up the A Level math questions!

An alternative way to solve this would be to factor the log2(x) out: 2log2(x) - log2(sqrt(x)) = 3 log2(x) * (2 - 0.5) = 3 (You can just think of moving the square root in front of the second log2) log2(x) = 3/1.5 x = 2^2 = 4

This is how HKDSE (university entrance exam in Hong Kong) ask a logarithm question [2023 Paper 1 Q.18]: Suppose that α, 7, β is a geometric sequence, where 1 (a) Express (log α base 7) in terms of (log β base 7) [3 marks] (b) If (log α base β), (log β base 7), (log β base α) is an arithmetic sequence, find the common difference of the arithmetic sequence. [5 marks]

@bprpmathbasics

Ай бұрын

Thanks for sharing. It’s 3am here now so I will take a look at this after I wake up 😆

@vedwargantiwar4610

Ай бұрын

I guess I will try it, alpha=A , beta=B a) 2-(logB base 7) b) common diffrence is 1.

@vedwargantiwar4610

Ай бұрын

Let me know if its wrong or right

@cyrusyeung8096

Ай бұрын

@@vedwargantiwar4610 Yes, you are correct.

@iwjrbf99

Ай бұрын

​@@cyrusyeung8096 dse add oil

I love you so much bprp! Thank you

The real magic of this channel is that the dry erase markers ALWAYS work.

I would love to see you talk through some AEA (advanced extension award) math questions as they cover the same material as a levels, but with way harder questions

you also can remove 2 before log2(x) by subtracting log2(x) on both sides, and it will give you the same answer) log2(x)-log2(sqrt(x))=3-log2(x) log2(x(sqrt(x))/x)=log2(8/x) sqrt(x)*x=8 x^3/2=2^3 x^3=2^6 x=4

I had basically the same correct solution, except I extracted the exponent from the second log first and combined like terms. 2*log_2(x) - (1/2)log_2(x) = 3, which goes to (3/2)log_2(x) = 3. Then, multiply both sides by 2/3 and raise 2 to the power of each side, and it's all good. I usually don't enjoy having to divide by the square root of anything.

This is one of those questions where you can just make a reasonable guess for what X is and substitute in. Plug in X=4 and you solve it in seconds. Of course, you'd get short shrift from the examiners without any working being shown.

Please make a video on solving VCE Exams from Australia (Mathematical Methods units 3/4), and Specialist Maths units 3/4). Rarely see it get covered in maths channels sadly!

cant believe my highschool days paid off , i can identify the log problems before i continue watching the video , i miss doing addmath problem

We can also Take x^3/2 in-front of Log right ? Then the 3 will cancel out and we get 4 as answer In the end part that is

my attempt at doing this: 2log_2(x) - log_2(sqrt(x)) = 3 2log_2(x) - (1/2)*log_2(x) = 3 (3/2)*log_2(x) = 3 log_2(x) = (2/3)*3 = 2 x = 2^2 = 4

(A level further math) Can you make video/explanation on question 8 (all) , on the paper ''Further Pure Mathematics 1 (Paper 3A) Friday 22nd May 2020'' Thanks

I see this question all the time! I even think it was an exam question for myself

would this be one of the easy test questions? it just looks like an ACT question (American standardized tests are very easy but try to trick you all the time)

I love the post-mortems!

@bprpmathbasics

Ай бұрын

?

Solution: First mistake: You cannot combine 2log₂(x) - log₂(√x) of any base. You first have to pull in the 2 as an exponent for the inner value, making it log₂(x²) - log₂(√x). So the actual result would be: log₂(x²/√x) = log₂(√x³) Second mistake: In the last step, he changed log₂(x) = 3 into x = 3², but it should be x = 2³. The actual result of the equation is: √x³ = 8 |² x³ = 64 |³√ x = 4

I solved it by 2log(2,x) - 1/2log(2,x) = (3/2)log(2,x) which is 1.5log(2,x) = 3 ..... / 1.5 log(2,x) = 2 2^2 = x x= 4

Looking at the thumbnail gave me an aneurysm

Alternatively 2log2(x)-log2(sqr(x) = 2log2(x)-1/2log2(x) =3/2log2(x) 3/2log2(x)=3 Log2(x)=2 x=4

You should try some of the harder questions from these papers, the two you've done are mostly designed to allow lower grade students to be able to pick up marks at the start and to allow the paper to be accessible

+ C 😅

You are claiming that the log had to have a coefficient of 1. This is false. Another way to look at it is to get the logarithm to be the same, so it is essentially the same variable. 2log x - 1/2 log x can easily be combined through the distributive property of multiplication, which means 3/2 log x = 3. log x = 2. x = 4. I was not sure how to denote base 2.

@tobybartels8426

Ай бұрын

It's traditional to use an underscore _ to indicate a subscript in a plain-text environment, so log_2 means log₂. Similarly, you can use a caret ^ to indicate a superscript, so x^2 means x². (You can also use Unicode characters, like I did here; search "Unicode subscript" to find somewhere to copy them from.)

@cx3622

Ай бұрын

It's not false. You're wrong. You can't combine log if the coefficient is one.

@ronaldking1054

Ай бұрын

@@cx3622 He's claiming you have only one rule to combine logarithms. That is false. Distributive property of multiplication allows the combination for logarithms with the same base and same value applied to the function. Logarithm is no different from any other function in this regard. This is a property of real numbers, and as such, you are claiming that logarithm is not a function of real numbers. That is false.

@cx3622

Ай бұрын

@@ronaldking1054 He didn't claim that. You made that up. Stop lying. You're wrong.

@ronaldking1054

Ай бұрын

@@cx3622 No, you did by stating what I said was wrong. Logarithm as a real number function ends up being able to apply the distributive property of multiplication to it. You do not need a coefficient of 1 to do it. The problem is that the student did not get to the same base logarithm, which means they were attempting to use another law incorrectly, but that specific law was not necessary to use at all.

Why is this on A Level math? I learnt logarithms which is tested for O levels

@moth5799

Ай бұрын

This is actually an AS maths question, I know because I'm in year 12 and just did this on a mock. A level questions are generally a bit harder than this, but the first few questions are still quite easy to allow the lower grade students to pick up some marks.

@nvapisces7011

Ай бұрын

@@moth5799 i would like to see a sample paper and see what is on it and isn't. I remember a lot of content being covered for my A Levels for H2 Mathematics. I can't say it's tough but definitely rigorous in terms of content

12th class Indian students can solve this question in 1 minute 🇮🇳🇮🇳