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Series logs question (1 Viewer)
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HeroicPandas
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is the answer the 11th term?
EDIT 1: heres what i did
This series is a geometric progression so the nth term is
Common ratio is -0.5 (i.e. 'r')
You aim is finding what 'n' is when
So... using nth term formula of geometric progression: (with r = -0.5, a = 8, n = ?)
^{n-1} \\ \\ \frac{1}{1024} = \left ( -1 \right )^{n-1} \left ( \frac{1}{2} \right )^{n-1} \\ \\$Take $log_{\frac{1}{2}}$ on both sides so we can get rid of the dirty base on the RHS to isolate 'n'\\ \\ $log_{\frac{1}{2}}\left ( \frac{1}{1024} \right ) = log_{\frac{1}{2}} \left [ (-1)^{n-1}\left ( \frac{1}{2} \right )^{n-1} \right ] \\ \\ log_{\frac{1}{2}}\left ( \frac{1}{1024} \right ) = log_{\frac{1}{2}}[(-1)^{n-1}] + log_{\frac{1}{2}}\left [ \left ( \frac{1}{2} \right )^{n-1} \right ] \\ \\ log_{\frac{1}{2}}\left ( \frac{1}{1024} \right ) = log_{\frac{1}{2}}\left [ \frac{(-1)^{n}}{-1} \right ] + log_{\frac{1}{2}}\left [ \left ( \frac{1}{2} \right )^{n-1} \right ])
 = log_{\frac{1}{2}}\left [ \frac{(-1)^{n}}{-1} \right ] + (n-1)log_{\frac{1}{2}}\left [ \left ( \frac{1}{2} \right ) \right ] \\ \\ log_{\frac{1}{2}}\left ( \frac{1}{1024} \right ) = log_{\frac{1}{2}}\left [ \frac{(-1)^{n}}{-1} \right ] + (n-1) )
Now recall natural domain of logarithms --------->
Looking on the RHS of the working,
^n}{-1} > 0)
So n must be odd--> if n is even, then (-1)n = -1, which makes^n}{-1} < 0 )
Hence n must be ODD
Since n is an odd integer, whatever it will be^n)
will be 1 (sentence A)
^{n} ] = \frac{ln(-(-1)^n)}{ln(0.5)} = \frac{0}{ln(0.5)})
<---because of sentence A
Continuing from working out,
 \\ \\ n = 1 + log_{\frac{1}{2}} \left [ \left ( \frac{1}{2} \right )^{10} \right ] \\ \\ n = 1 + 10 \\ n = 11)
Sorry for this weird working out, hope u understand (havent done math in a while, so i might miss something)
EDIT 2: maybe changing
into 
would have made it easier...
EDIT 1: heres what i did
This series is a geometric progression so the nth term is
Common ratio is -0.5 (i.e. 'r')
You aim is finding what 'n' is when
So... using nth term formula of geometric progression: (with r = -0.5, a = 8, n = ?)
Now recall natural domain of logarithms --------->
Looking on the RHS of the working,
So n must be odd--> if n is even, then (-1)n = -1, which makes
Hence n must be ODD
Since n is an odd integer, whatever it will be
Continuing from working out,
Sorry for this weird working out, hope u understand (havent done math in a while, so i might miss something)
EDIT 2: maybe changing
Last edited:
RealiseNothing
what is that?It is Cowpea
So there are 7 fraction terms, plus 1, 2, 4, 8.
so 11th term
anomalousdecay
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is the answer the 11th term?
EDIT 1: heres what i did
This series is a geometric progression so the nth term is
Common ratio is -0.5 (i.e. 'r')
You aim is finding what 'n' is when
So... using nth term formula of geometric progression: (with r = -0.5, a = 8, n = ?)
^{n-1} \\ \\ \frac{1}{1024} = \left ( -1 \right )^{n-1} \left ( \frac{1}{2} \right )^{n-1} )
Just use this line. Do this:
Use the calculator obviously through log laws re-arrangement. It can't be negative as this is 2-unit
Then you get n=11.
Simple.
Last edited:
HeroicPandas
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Just use this line. Do this:
Use the calculator obviously through log laws re-arrangement. It can't be negative as this is 2-unit
Then you get n=11.
Simple.
why cant there be a negative?
anomalousdecay
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There is no log negative in even 4-unit, let alone 3-unit.which is impossible because ln(-blah) is undefined
why cant there be a negative?
Also, when integrating in 4-unit, you should use the absolute value signs for log.
The re-arrangement used in your first post was much more suitable than this, as it showed the splitting of the negative one.
By using logic, it also ties in with what you said before, how n is odd.
Also, if the term was negative, then n would have to be positive.
There are no marks assigned for these statements though in HSC.
Btw, did an all-nighter, but fell asleep at 5am for 5 hours haha. Your sig
Last edited:
HeroicPandas
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aww never knew that hahaThere is no log negative in even 4-unit, let alone 3-unit.
Also, when integrating in 4-unit, you should use the absolute value signs for log.
The re-arrangement used in your first post was much more suitable than this, as it showed the splitting of the negative one.
By using logic, it also ties in with what you said before, how n is odd.
Also, if the term was negative, then n would have to be positive.
There are no marks assigned for these statements though in HSC.
Btw, did an all-nighter, but fell asleep at 5am for 5 hours haha. Your sig
eh..all-nighter lol