I have the expressions for the nth term of various sequences. I'm supposed to write them in a "p/q" fraction form, but i have no idea how to do this with logarithmic expressions.

log2^n(8)

log3^n(81)

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- Feb 15th 2010, 04:55 PMorganismalWriting an expression in fraction form
I have the expressions for the nth term of various sequences. I'm supposed to write them in a "p/q" fraction form, but i have no idea how to do this with logarithmic expressions.

log2^n(8)

log3^n(81) - Feb 15th 2010, 06:26 PMtonio
- Feb 15th 2010, 07:47 PMorganismal
i'm sorry if im misunderstanding what you're saying, but it's not an equation..?

i was given a sequence:

$\displaystyle

\log_{2}8, \log_{4}8, \log_{8}8, \log_{16}8, \log_{32}8

$

i was supposed to write down the next two terms:

$\displaystyle

\log_{64}8, \log_{128}8

$

and find an expression fo the nth term of each sequence

$\displaystyle

\log_{2^n}8

$

which i am hoping is correct. then it says to write your expression in the form p/q, meaning a fraction (or I suppose a whole number if it works out that way).

i tried writing it in exponental form and got to $\displaystyle (2^n)^x=8$ but that doesn't make any sense to me

edit: nevermind, i'm sorry, i think i just understood what it was was asking for and your explanation. thank you very much! - Feb 15th 2010, 08:04 PMProve It
$\displaystyle \log_2{8} = \log_{2^1}{8} = \log_{2^1}{(2^1)^\frac{3}{1}} = \frac{3}{1}$

$\displaystyle \log_4{8} = \log_{2^2}{8} = \log_{2^2}{(2^2)^{\frac{3}{2}}} = \frac{3}{2}$

$\displaystyle \log_8{8} = \log_{2^3}{8} = \log_{2^3}{(2^3)^{\frac{3}{3}}} = \frac{3}{3}$

$\displaystyle \log_8{8} = \log_{2^4}{8} = \log_{2^4}{(2^4)^{\frac{3}{4}}} = \frac{3}{4}$

$\displaystyle \log_{16}{8} = \log_{2^5}{8} = \log_{2^5}{(2^5)^{\frac{3}{5}}} = \frac{3}{5}$.

I think you can see that:

$\displaystyle \log_{2^n}{8} = \frac{3}{n}$.