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expontial model

Hi all
A researcher discoversthat between 2008 and 2010, the number of messages sent on Twitter (called tweets) increased exponentially. She investigates the number of tweets from apopular author, and finds that they can be modelledby the equation u = 20×(1.8)t, where u is the total number of tweets sent, and t is the number of months since the author became a Twitter user, where 0 t 8.1. Use the method shown in Unit 13, Subsection 5.2 to find the month in which the exponential model predicts that the total number of tweets sent will first reach 500
i did it like this u = log(10) * 25 log(10) * 1.8 = 45 but 45 doesnt seem right if you put 45 in the model u= 20*(1.8)*45 = 1620 which is far more than 500. but if i put 14 in the model u = 20- *(1.8) * 14 = 504 tweets somonth 14 seems more right put the logs dont make sense .Any help with logs please.Eggie
Original post by eggie
Hi all
A researcher discoversthat between 2008 and 2010, the number of messages sent on Twitter (called tweets) increased exponentially. She investigates the number of tweets from apopular author, and finds that they can be modelledby the equation u = 20×(1.8)t, where u is the total number of tweets sent, and t is the number of months since the author became a Twitter user, where 0 t 8.1. Use the method shown in Unit 13, Subsection 5.2 to find the month in which the exponential model predicts that the total number of tweets sent will first reach 500
i did it like this u = log(10) * 25 log(10) * 1.8 = 45 but 45 doesnt seem right if you put 45 in the model u= 20*(1.8)*45 = 1620 which is far more than 500. but if i put 14 in the model u = 20- *(1.8) * 14 = 504 tweets somonth 14 seems more right put the logs dont make sense .Any help with logs please.Eggie


Do you mean u=201.8tu=20 \cdot 1.8^t?

So then u=500u=500 and 500=201.8t500=20 \cdot 1.8^t

Take logs of both sides and use some log rules:
log(500)=log(201.8t)=log(20)+log(1.8t)log(500)=log(20 \cdot 1.8^t)=log(20)+log(1.8^t)

Im not sure where you are pulling log(10) from.
Reply 2
Original post by RDKGames
Do you mean u=201.8tu=20 \cdot 1.8^t?

So then u=500u=500 and 500=201.8t500=20 \cdot 1.8^t

Take logs of both sides and use some log rules:
log(500)=log(201.8t)=log(20)+log(1.8t)log(500)=log(20 \cdot 1.8^t)=log(20)+log(1.8^t)

Im not sure where you are pulling log(10) from.


Hi RDk
It gives base log(10) in the examples so thats what im using to answer the question. It asks for month 1 , month 2 month 3 and month 8 which are
month 1 = u = 20 * 1.8 * 1 = 36
month 2 = u = 20* 1.8 * 2 = 72
month 3 = u = 20* 1.8 * 3 = 108
month 8 = u = 20 * 1.8 * 8 = 288

so the models scale factor is 100% so month 7 = u = 20* 1.8 * 7 = 252 so month 14 =504 tweets as i said in the example.

u = 20*(1.8)^t which month tweets reach 500
I am dividing 20 by 500 to get 25 and putting 25 in the logs but taking the log20 and log1.8^t still cant get 14 tried different scenarios but still stuck .

any extra help appreciated
Original post by eggie
Hi RDk
It gives base log(10) in the examples so thats what im using to answer the question. It asks for month 1 , month 2 month 3 and month 8 which are
month 1 = u = 20 * 1.8 * 1 = 36
month 2 = u = 20* 1.8 * 2 = 72
month 3 = u = 20* 1.8 * 3 = 108
month 8 = u = 20 * 1.8 * 8 = 288

so the models scale factor is 100% so month 7 = u = 20* 1.8 * 7 = 252 so month 14 =504 tweets as i said in the example.

u = 20*(1.8)^t which month tweets reach 500



tt is bounded between 0 and 8.1. It CANNOT be 14 months. The equation is u=201.8tu=20\cdot 1.8^t and NOT u=201.8tu=20\cdot 1.8t otherwise it will never reach 500 within 8 months.

Also, what do you mean by:
I am dividing 20 by 500 to get 25 and putting 25 in the logs but taking the log20 and log1.8^t still cant get 14 tried different scenarios but still stuck .


First of all, 20 divided by 500 is NOT 25. Secondly, that's the wrong order of operation. Lastly, putting 25 in the logs doesn't make sense, I think you meant taking the log of 500.

You are looking for the time when u=500. So plug it into your equation and solve for t.

500=201.8t500=20 \cdot 1.8^t
25=1.8t\rightarrow 25=1.8^t and now solve it using logs.
(edited 7 years ago)
Reply 4
Original post by RDKGames
tt is bounded between 0 and 8.1. It CANNOT be 14 months. The equation is u=201.8tu=20\cdot 1.8^t and NOT u=201.8tu=20\cdot 1.8t otherwise it will never reach 500 within 8 months.

Also, what do you mean by:


First of all, 20 divided by 500 is NOT 25. Secondly, that's the wrong order of operation. Lastly, putting 25 in the logs doesn't make sense, I think you meant taking the log of 500.

You are looking for the time when u=500. So plug it into your equation and solve for t.

500=201.8t500=20 \cdot 1.8^t
25=1.8t\rightarrow 25=1.8^t and now solve it using logs.


Hi RDK

I charted a graph using the equation and got month 5.47 when tweets reach 500

i did it this way using log base 10 and get the same answer

u = log(10)*25 / log(10) * 1.8 = 5.47 graph gives the same answer when i plot
the point 500. Now only one thing i cant work out is tweets for month 8 but i think it is 2204. Is that right can u check my answers fot tweets

month 1 = 36
month2 = 64.8 or 65
month3 = 116.6 or 117
month8 = 2204

are these right please

thanks

Eggie
Original post by eggie
Hi RDK

I charted a graph using the equation and got month 5.47 when tweets reach 500

i did it this way using log base 10 and get the same answer

u = log(10)*25 / log(10) * 1.8 = 5.47 graph gives the same answer when i plot
the point 500. Now only one thing i cant work out is tweets for month 8 but i think it is 2204. Is that right can u check my answers fot tweets

month 1 = 36
month2 = 64.8 or 65
month3 = 116.6 or 117
month8 = 2204

are these right please

thanks

Eggie


Yes those are correct.
Reply 6
Original post by RDKGames
Yes those are correct.


Hi RDk

Thanks for the help only last thing left and finish my End of module assessment
and finish the course is the scale factor how do i work out the scale factor of the
expontenial model and percentage increase in total number tweets sent

thanks
Eggie

found the scale factor which 1.80 or 80% but how do i find the percentage increase in total number of tweets sent
from 36 tweets to 2204 tweets so how do i find the percentage increase please.
(edited 7 years ago)
Original post by eggie
Hi RDk

Thanks for the help only last thing left and finish my End of module assessment
and finish the course is the scale factor how do i work out the scale factor of the
expontenial model and percentage increase in total number tweets sent

thanks
Eggie

found the scale factor which 1.80 or 80% but how do i find the percentage increase in total number of tweets sent
from 36 tweets to 2204 tweets so how do i find the percentage increase please.


(2204-36)/36 * 100% = 6022%.

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