# Thread: Logarthim - Finding expression - URGENT PLZZ

1. ## 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!!!!!!

2. 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 $\displaystyle a_n = \log_{2^n}8$ for $\displaystyle n = 1,~2,~3, \cdots$ ?

3. 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 $\displaystyle a_n = \log_{2^n}8$ for $\displaystyle n = 1,~2,~3, \cdots$ ?
$\displaystyle a_{n}=\log_{2^{n}}(8) \Rightarrow 2^{na_{n}}=2^{3}$
....................$\displaystyle \Rightarrow 2^{na_{n}-3}=1$
....................$\displaystyle \Rightarrow na_{n}-3=0$
....................$\displaystyle \Rightarrow a_{n}={\color{red}{\frac{3}{n}}}.$

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

I dont think that this is a geometric sequence.
Or if it is so, I have a big mistake.

4. Originally Posted by bkarpuz
$\displaystyle a_{n}=\log_{2^{n}}(8) \Rightarrow 2^{na_{n}}=2^{3}$
....................$\displaystyle \Rightarrow 2^{na_{n}-3}=1$
....................$\displaystyle \Rightarrow na_{n}-3=0$
....................$\displaystyle \Rightarrow a_{n}=-\frac{3}{n}.$ *

[snip]
Correction to line *: $\displaystyle \Rightarrow a_{n}=\frac{3}{n}.$

5. 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 ....?

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

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

7. 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

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.