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# Using Log base 10 to find gravity watch

1. (Original post by student403)
ignore my above post

your intercept of 0.302 is correct. zacken watch out intercept of -0.2 is where x = approx (-1), not where x = 0

see the x axis labels

anyway the second reason you were not getting the correct value of g is because of the base.

Now that you have 0.302 as your intercept, you need to be taking 10 to the 0.302, not e to the 0.302

so with 10^0,302 = 2pi/sqrtg

you can solve to get g correct to 2 s.f.
student 403 is god tier.
2. (Original post by Jyashi)
student 403 is god tier.
Zacken is way above that. He's just tired!

But glad you get it now
3. (Original post by Jyashi)
student 403 is god tier.
(Original post by Student403)
Zacken is way above that. He's just tired!

But glad you get it now
Nope, definitely agree with Jyashi!
4. (Original post by Zacken)
Nope, definitely agree with Jyashi!
PRSOM
5. (Original post by Zacken)
Nope, definitely agree with Jyashi!
I would say you guys are both god tier. I would like to become god tier myself one day.

Looking back now i see the equation that Zacken posted which i cant understand why square root g became the denominator of 2pi. Can you guys explain the process of that? I reworked the equation again and got the below answer.

6. (Original post by Jyashi)
I would say you guys are both god tier. I would like to become god tier myself one day.

Looking back now i see the equation that Zacken posted which i cant understand why square root g became the denominator of 2pi. Can you guys explain the process of that? I reworked the equation again and got the below answer.

Power rule, 1/2 log g = log sqrt(g)
7. (Original post by Jyashi)
I would say you guys are both god tier. I would like to become god tier myself one day.

Looking back now i see the equation that Zacken posted which i cant understand why square root g became the denominator of 2pi. Can you guys explain the process of that? I reworked the equation again and got the below answer.

Careful in your working. Remember if you're going to make it a -log(P) you can't take the exponent to be negative as well. It's one or the other in this case (referring to your second line)

No need to take the half out of the log

Since you have

LogT = Log2PI + 0.5LogL - Log(g^0.5)

Now using the log rule of LogA - LogB = Log(A/B)

You can get

LogT = Log(2PI/g^0.5) + 0.5LogL
8. (Original post by Student403)
Careful in your working. Remember if you're going to make it a -log(P) you can't take the exponent to be negative as well. It's one or the other in this case (referring to your second line)

No need to take the half out of the log

Since you have

LogT = Log2PI + 0.5LogL - Log(g^0.5)

Now using the log rule of LogA - LogB = Log(A/B)

You can get

LogT = Log(2PI/g^0.5) + 0.5LogL
You should seriously consider a career in teaching. Got that perfectly.
9. (Original post by Jyashi)
You should seriously consider a career in teaching. Got that perfectly.
My pleasure and thank you

You should see Zacken's maths help though.. He gets anything pure.. Literally everything

If you ever need pure maths help, he's your man!
10. (Original post by Student403)
My pleasure and thank you

You should see Zacken's maths help though.. He gets anything pure.. Literally everything

If you ever need pure maths help, he's your man!
Thanks both of you have been awesome help for me. I think this wraps up my doubts for this question.
11. (Original post by Jyashi)
Thanks both of you have been awesome help for me. I think this wraps up my doubts for this question.
Nice going - we got there! Come back any time if you need help

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