Logarthim - Finding expression - URGENT PLZZ

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September 28th 2008, 11:53 PM
GoneCrazy

Logarthim - Finding expression - URGENT PLZZ

log[base2]8, log[base4]8, log[base8]8, log[base18, log[base32]8, log[base64]8,

I am given this sequence and i need to find an expression for the nth term in form of p/q, where p and q belongs to z(integer).

i guess it has to do with the bases
like log[base2]8
= log8/log2
= log8 - log2
= ??????
then how to find an expression

I tried the following:

1) since the bases form a geometric progression
i found the common ratio
which is r =Un/Un-1
ex: 8/4 = 2

2) The nth term formula for a geometric progression
is Un= ar^(n-1)
where a = 1st term, which is Log(base2)8
r=common ratio, which is 2
n=the rank

so the expression is
Un= Log(base2)8(2)^n-1
However its not quite correct
beacuse if we tested it, when n=7
to complete the sequence
the value should be log(base128)8
NOT Log(base2)8 x 128
I hope you get where i am going

I really appreciate ypu help
THANK YOU!!!!!!
September 28th 2008, 11:58 PM
Jhevon

Quote:

Originally Posted by GoneCrazy

log[base2]8, log[base4]8, log[base8]8, log[base18, log[base32]8, log[base64]8,

I am given this sequence and i need to find an expression for the nth term in form of p/q, where p and q belongs to z(integer).

i guess it has to do with the bases
like log[base2]8
= log8/log2
= log8 - log2
= ??????
then how to find an expression

I tried the following:

1) since the bases form a geometric progression
i found the common ratio
which is r =Un/Un-1
ex: 8/4 = 2

2) The nth term formula for a geometric progression
is Un= ar^(n-1)
where a = 1st term, which is Log(base2)8
r=common ratio, which is 2
n=the rank

so the expression is
Un= Log(base2)8(2)^n-1
However its not quite correct
beacuse if we tested it, when n=7
to complete the sequence
the value should be log(base128)8
NOT Log(base2)8 x 128
I hope you get where i am going

I really appreciate ypu help
THANK YOU!!!!!!

you said the base was a geometric progression. why did you not then make the base the geometric progression? i have trouble seeing your logic. would it not rather be $a_n = \log_{2^n}8$ for $n = 1,~2,~3, \cdots$ ?
September 29th 2008, 01:44 AM
bkarpuz

Quote:

Originally Posted by Jhevon

you said the base was a geometric progression. why did you not then make the base the geometric progression? i have trouble seeing your logic. would it not rather be $a_n = \log_{2^n}8$ for $n = 1,~2,~3, \cdots$ ?

$a_{n}=\log_{2^{n}}(8) \Rightarrow 2^{na_{n}}=2^{3}$
.................... $\Rightarrow 2^{na_{n}-3}=1$
.................... $\Rightarrow na_{n}-3=0$
.................... $\Rightarrow a_{n}={\color{red}{\frac{3}{n}}}.$

$a_{n}={\color{red}{\frac{n-1}{n}}}{\color{red}{\frac{3}{n-1}}}$
..._ $={\color{red}{\frac{n-1}{n}a_{n}}}$

I dont think that this is a geometric sequence. (Thinking)
Or if it is so, I have a big mistake. (Doh)
September 29th 2008, 02:46 AM
mr fantastic

Quote:

Originally Posted by bkarpuz

$a_{n}=\log_{2^{n}}(8) \Rightarrow 2^{na_{n}}=2^{3}$
.................... $\Rightarrow 2^{na_{n}-3}=1$
.................... $\Rightarrow na_{n}-3=0$
.................... $\Rightarrow a_{n}=-\frac{3}{n}.$ *

[snip]

Correction to line *: $\Rightarrow a_{n}=\frac{3}{n}.$
September 29th 2008, 02:48 AM
mr fantastic

Quote:

Originally Posted by GoneCrazy

log[base2]8, log[base4]8, log[base8]8, log[base18, log[base32]8, log[base64]8,

I am given this sequence and i need to find an expression for the nth term in form of p/q, where p and q belongs to z(integer).
[snip]

Regarding the red: You mean log[base16]8, right ....?
September 29th 2008, 08:36 AM
GoneCrazy

Yeh it has nothing to do with geometric progression -my mistake
i didn't take enough time to think it over before asking (Itwasntme)

oooh yeh...i meant 16 not 18 - sorry

I really didn't get your way - Both bkarpuz and Mr.Fantastic,
if you can explain it more plzzzz

---------------------------------------------------

Here is my way, i figured out 2 ways:

2,4,8,16,32,64 can be expressed as 2 raised to a certain power
Then using these two basic rules, each value in the squence can be simplified:
1) Log (baseB)A = Log A /Log B
2) Log a^2 = 2Loga

example: Log (base4 )8

= Log 8
Log 4

= Log 2^3
Log 2^2

3 Log 2
2 Log 2

= 3
2

The same applies for the rest
making the following pattern:
3/1 , 3,2 , 3/3, 3/4, 3/5, 3/6

Therefore, the nth-term is going to be 3/n

--------------------------------------------

2) The sequence is in the form of log[base 2^n] 8

By using the base rule, we can further simplify it beacome in the form of p/q

log[base 2^n] 8
= log[base 2] 8 / log [base 2]2^n
= log[base 2] 2^3 / log [base 2]2^n

After crossing out the similar logs you are left with 3/n
----------------------------------
I have anthor part of the question if anyone is intrested
September 29th 2008, 12:55 PM
mr fantastic

Quote:

Originally Posted by GoneCrazy

Yeh it has nothing to do with geometric progression -my mistake
i didn't take enough time to think it over before asking (Itwasntme)

oooh yeh...i meant 16 not 18 - sorry

I really didn't get your way - Both bkarpuz and Mr.Fantastic,
if you can explain it more plzzzz

---------------------------------------------------

Here is my way, i figured out 2 ways:

2,4,8,16,32,64 can be expressed as 2 raised to a certain power
Then using these two basic rules, each value in the squence can be simplified:
1) Log (baseB)A = Log A /Log B
2) Log a^2 = 2Loga

example: Log (base4 )8

= Log 8
Log 4

= Log 2^3
Log 2^2

3 Log 2
2 Log 2

= 3
2

The same applies for the rest
making the following pattern:
3/1 , 3,2 , 3/3, 3/4, 3/5, 3/6

Therefore, the nth-term is going to be 3/n

--------------------------------------------

2) The sequence is in the form of log[base 2^n] 8

By using the base rule, we can further simplify it beacome in the form of p/q

log[base 2^n] 8
= log[base 2] 8 / log [base 2]2^n
= log[base 2] 2^3 / log [base 2]2^n

After crossing out the similar logs you are left with 3/n
----------------------------------
I have anthor part of the question if anyone is intrested

If you have more questions related to this one then you should certainly feel free to ask them.