Usually you dont write decimals for standard form so instead of ....x10^26.2 just write in form ....x10^26

Original post by mqb2766

The number 5.21 is correct, though the setting up is confusing.

The actual algebra appears ok.

•

What are the Ms so 5M and 5.21M, why equals ... M is a variable with values 5 and 5.21

•

Why 2E=1900 then 2E=....

•

in part i) why 1989 = 5M?

The 1989=5M and 1900=2E was just reference from the question asked…

So, in 1989 the magnitude was 5m and second part in 1900 energy release was 2E from 1989…

Do you think I should remove this to remove any confusion?

Original post by mosaurlodon

Usually you dont write decimals for standard form so instead of ....x10^26.2 just write in form ....x10^26

So, it’s okay to remove the .2. Is that the only thing that appears wrong? What if they’re expecting the actual number? 2.757714155x₁₀²⁶

Original post by KingRich

The 1989=5M and 1900=2E was just reference from the question asked…

So, in 1989 the magnitude was 5m and second part in 1900 energy release was 2E from 1989…

Do you think I should remove this to remove any confusion?

So, in 1989 the magnitude was 5m and second part in 1900 energy release was 2E from 1989…

Do you think I should remove this to remove any confusion?

E and M are variables defined in the question, yet you seem to be using them as shorthand for energy and magnitude which is not standard, to say the least. It doesnt make sense what youve written. You could say something like

•

in 1989, M=5 and E=2.7*10^26 J

•

in 1900, E= 5.4*10^26, we need to determine M.

(edited 1 month ago)

Well your answer is not wrong technically, ive just never seen standard form written like that in any answer - I guess im just telling you to stop being weird

Original post by mqb2766

E and M are variables defined in the question, yet you seem to be using them as shorthand for energy and magnitude which is not standard, to say the least. It doesnt make sense what youve written. You could say something like

Note the scientific notation mentioned above and yesterday.

•

in 1989, M=5 and E=2.7*10^26 J

•

in 1900, E= 5.4*10^26, we need to determine M.

I see what you mean, using equal signs could be confusing for the person checking my work.

So, from what I recall yesterday. I had a little trouble trying to understand.

Let’s say, I have this. E=1.74 X 10²⁶ X 10^0.2

How do I find that solution 2.7 x 10²⁶ without using the calculator?

Original post by KingRich

I see what you mean, using equal signs could be confusing for the person checking my work.

So, from what I recall yesterday. I had a little trouble trying to understand.

Let’s say, I have this. E=1.74 X 10²⁶ X 10^0.2

How do I find that solution 2.7 x 10²⁶ without using the calculator?

So, from what I recall yesterday. I had a little trouble trying to understand.

Let’s say, I have this. E=1.74 X 10²⁶ X 10^0.2

How do I find that solution 2.7 x 10²⁶ without using the calculator?

You need a calculator to work out 10^0.2 or log_10(2) for instance.

It is confusing the way you use equals and reuse M, E. The equals sign was introduced by Robert Recorde in the 1500s

https://en.wikipedia.org/wiki/Equals_sign

and has two equal horizontal lines to denote the fact that each side of it (length of each line) has equal value. Youre not using it in that way.

(edited 1 month ago)

Original post by KingRich

Ah, right, okay.

I suppose because none of my previous tutors has ever said anything. They'd just looked at my main calculations and ignored any side notes that I made, I never really thought about it.

I suppose because none of my previous tutors has ever said anything. They'd just looked at my main calculations and ignored any side notes that I made, I never really thought about it.

Your calcs are ok here, though as per yesterday you could simplify the final expression a bit as

M = (log(2) + 7.2)/1.44 = 5 + log(2)/1.44

as you have log(2*10^26.2) which can be transformed.

(edited 1 month ago)

Original post by mqb2766

Youre calcs are ok here, though as per yesterday you could simplify the final expression abit as

M = (log(2) + 7.2)/1.44 = 5 + log(2)/1.44

as you have log(2*10^26.2) which can be transformed.

M = (log(2) + 7.2)/1.44 = 5 + log(2)/1.44

as you have log(2*10^26.2) which can be transformed.

Do you mean part cii?

Original post by KingRich

Do you mean part cii?

Yes, you something like

log(4*10^26.2/2) = log(2*10^26.2) = log(2) + 26.2 ....

Original post by mqb2766

Yes, you something like

log(4*10^26.2/2) = log(2*10^26.2) = log(2) + 26.2 ....

log(4*10^26.2/2) = log(2*10^26.2) = log(2) + 26.2 ....

I was just wondering, for part ci, do you think they may have been looking for me to apply log in order find 26? Or, would it have to hint in some degree by I don’t know, saying simply as far as possible.

I’m concerned as this might be what was intended?

Original post by KingRich

I was just wondering, for part ci, do you think they may have been looking for me to apply log in order find 26? Or, would it have to hint in some degree by I don’t know, saying simply as far as possible.

I’m concerned as this might be what was intended?

I’m concerned as this might be what was intended?

Best guess from me, they would be expecting the answer 2.76*10^26 for i). For ii) they would be expecting the (algebraic) answer of 5+log_10(2)/1.44, as its simple and can be easily interpreted, which then evaluates to 5.21.

However, ask your tutor if necessary, but you may be overthinking it.

Note the "simplest" way to do the second part is to note

5 -> E, so

E = 1.74*10^19*10^(1.44*5)

M -> 2E, so

2E = 1.74*10^19*10^(1.44*M)

Divide the second by the first you get

2 = 10^(1.44*(M-5))

take log base 10 and rearrange.

(edited 1 month ago)

- If formula E=(1.74x10^19 x 10^1.44M)
- Buffer calculation help
- Biology maths
- STEP maths I, II, III 1991 solutions
- STEP maths I, II, III 1990 solutions
- The heat released when 1.00 g of ethanol (Mr = 46.0) undergoes complete combustion is
- Chemistry A Level Titrations Help
- how to solve this logathrithm problem?
- STEP Maths I, II, III 1993 Solutions
- AQA A Level Chemistry Paper 3 20th June 2018 Unofficial Markscheme
- Vectors help?
- Physics A G482 EWP Unofficial mark Scheme
- Edexcel A-level Further Mathematics Paper 2 (9FM0 02) - 5th June 2023 [Exam Chat]
- STEP Maths I, II, III 1996 Solutions
- Edexcel A-level Further Mathematics Paper 2 (9FM0 02) - 3rd June 2024 [Exam Chat]
- STEP 2016 Solutions
- Edexcel IGCSE 24th May 2018 Paper 3H Unofficial mark Sceheme
- Alevel chemistry kw
- physics question
- AQA GCSE Mathematics Paper 2 Higher (8300/2H) - 3rd June 2024 [Exam Chat]

Last reply 1 week ago

STEP 2 in 2024: Sharing Your Story! [PLUS WITH SOME SOLUTIONS AND PREDICTION]Maths

18

80

Last reply 1 week ago

A level maths paper 2 (pure and statistics) and paper 3 (pure and mechanics) ocrMaths

4

6

Last reply 1 week ago

STEP 2 in 2024: Sharing Your Story! [PLUS WITH SOME SOLUTIONS AND PREDICTION]Maths

18

80

Last reply 1 week ago

A level maths paper 2 (pure and statistics) and paper 3 (pure and mechanics) ocrMaths

4

6