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geometric progession help

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no

3 * 5^x = 150

5^x = 50

log5^x = log50

xlog5 = log50

x = log50/log5


Can I ask if you are self teaching this material ... if not, there seems to be a lot that you have not understood from lessons
Original post by dongonaeatu
is the sum to infinity 16.6


as my post said ... NO
Reply 22
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
hey man, solve 3*5^x=150 giving it to 4dp

is it; log3*5^x=log150

5^x=log150/log3

5^x=4.560876795 then i dont know how to get the x on its own


3×5x=150  ⟹  5x=50 \displaystyle 3 \times 5^{x} = 150 \implies 5^{x} = 50

Now take log of both sides and remember, logba=alogb \displaystyle logb^a = alogb
Reply 23
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
3×5x=150  ⟹  5x=50 \displaystyle 3 \times 5^{x} = 150 \implies 5^{x} = 50

Now take log of both sides and remember, logba=alogb \displaystyle logb^a = alogb


oh i see so you get rid of the 3* so its 5^x=50 then do i do

log5^x=log50
Reply 24
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
oh i see so you get rid of the 3* so its 5^x=50 then do i do

log5^x=log50


Yes
Reply 25
Avatar for dongonaeatu
dongonaeatu
OP
Original post by TenOfThem
no

3 * 5^x = 150

5^x = 50

log5^x = log50

xlog5 = log50

x = log50/log5


Can I ask if you are self teaching this material ... if not, there seems to be a lot that you have not understood from lessons


thanks that was helpful, is the answer 2.4307? and no i do AS maths at college
Reply 26
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
oh i see so you get rid of the 3* so its 5^x=50 then do i do

log5^x=log50


By the way, you need to learn the log chapter again. Going through your book will be more helpful than asking questions here.

I am quite sure if you learn the chapter well, then you should be able to answer most of these questions.
Reply 27
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
thanks that was helpful, is the answer 2.4307? and no i do AS maths at college


Yes
Reply 28
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
By the way, you need to learn the log chapter again. Going through your book will be more helpful than asking questions here.

I am quite sure if you learn the chapter well, then you should be able to answer most of these questions.


thanks man i will do that
Reply 29
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
Yes


The first term of a geometric progression is 5 and the third term is 10
i. determine the two possible values for the common ratio [2marks]

so a=5 and ar is the second term so ar^2=10

so 5(r)^2=10

r^2=5 so r=square root of 5 so thats 2.236067977 or -2.236067977

is this right
(edited 12 years ago)
Reply 30
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
Yes


sorry that was wrong

i think its: ar^2=10
5(r)^2=10
so then i divide by 5 NOT minus 5 like i did last time becauses it's 5 times

so r^2=2
so r= square root of 2 which equals 1.414213562 or -1.414213562 please tell me this is correct
Reply 31
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
sorry that was wrong

i think its: ar^2=10
5(r)^2=10
so then i divide by 5 NOT minus 5 like i did last time becauses it's 5 times

so r^2=2
so r= square root of 2 which equals 1.414213562 or -1.414213562 please tell me this is correct


It is correct :wink:
Reply 32
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
It is correct :wink:


yay!!

ii. Using the largest of these two values for the common ratio find the first term to exceed 5000. [3 marks]

so i am using the positive 1.414213562 but i really dont know how to tackle this type of question
Reply 33
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
yay!!

ii. Using the largest of these two values for the common ratio find the first term to exceed 5000. [3 marks]

so i am using the positive 1.414213562 but i really dont know how to tackle this type of question


You know that nth term=arn−1 = ar^{n-1}

So here, 5000=5rn−1 5000=5r^{n-1}

Find the value of 'n'.
Reply 34
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
You know that nth term=arn−1 = ar^{n-1}

So here, 5000=5rn−1 5000=5r^{n-1}

Find the value of 'n'.


so 5000= 5(1.414213562)^n-1

5000=7.07106701^n-1

how do i get n
Reply 35
Avatar for Joshmeid
Joshmeid
5
Original post by dongonaeatu
so 5000= 5(1.414213562)^n-1

5000=7.07106701^n-1

how do i get n


Take the log of both sides :smile:

Also to make it easier, keep it in root form :biggrin:

As in, instead of 1.414213562, use 2\sqrt 2
(edited 12 years ago)
Reply 36
Avatar for dongonaeatu
dongonaeatu
OP
Original post by Joshmeid
Take the log of both sides :smile:


this isnt a log question its geometric progression
Reply 37
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
so 5000= 5(1.414213562)^n-1

5000=7.07106701^n-1

how do i get n


You really need to go through the chapter again.

5000=5(2)n−1  ⟹  1000=(2)n−1  ⟹  log1000=log(2)n−1  ⟹  log1000=(n−1)log2  ⟹  log1000=nlog2−log2  ⟹  log1000+log2log2=n  ⟹  n=20.9315 \displaystyle 5000=5(\sqrt2)^{n-1} \implies 1000=(\sqrt2)^{n-1} \implies log1000=log(\sqrt2)^{n-1} \implies log1000=(n-1)log\sqrt2 \implies log1000=nlog\sqrt2 - log\sqrt2 \implies \frac{log1000+log\sqrt2}{log \sqrt2}=n \implies n=20.9315

The answer is n=21 \boxed{n=21}
Reply 38
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
You really need to go through the chapter again.

5000=5(2)n−1  ⟹  1000=(2)n−1  ⟹  log1000=log(2)n−1  ⟹  log1000=(n−1)log2  ⟹  log1000=nlog2−log2  ⟹  log1000+log2log2=n  ⟹  n=20.9315 \displaystyle 5000=5(\sqrt2)^{n-1} \implies 1000=(\sqrt2)^{n-1} \implies log1000=log(\sqrt2)^{n-1} \implies log1000=(n-1)log\sqrt2 \implies log1000=nlog\sqrt2 - log\sqrt2 \implies \frac{log1000+log\sqrt2}{log \sqrt2}=n \implies n=20.9315

The answer is n=21 \boxed{n=21}


thats insane, and i never knew logs were used in geometric progressions; how did you know to use logs
Reply 39
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
thats insane, and i never knew logs were used in geometric progressions; how did you know to use logs


The only way to solve this was to use logs, as said before, you need to go through your book again.

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