# Thread: Writing an expression in fraction form

1. ## Writing 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)

2. Originally Posted by organismal
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)

If, for example , you meant $\displaystyle \log_{2^n}8=\frac{3}{n}$ , and this follows at once from the very definition of logarithms. Check it and understand it.

Tonio

3. 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!

4. Originally Posted by organismal
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
$\displaystyle \log_2{8} = \log_{2^1}{8} = \log_{2^1}{(2^1)^\frac{3}{1}} = \frac{3}{1}$

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

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

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

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

I think you can see that:

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