# Finding nth term

• Oct 25th 2008, 07:16 AM
vale01
Finding nth term
Sequence: log(subscript)2) 8, log(subscript)4) 8, log(subscript)8) 8, log(subscript)16) 8, log(subscript)32) 8, log(subscript)64) 8…..
How do I find an expression for the nth term of the sequence in the form p/q, where p and q are integers.
Have an awesome day!!
• Oct 25th 2008, 07:26 AM
flyingsquirrel
Hi,
Quote:

Originally Posted by vale01
Sequence: log(subscript)2) 8, log(subscript)4) 8, log(subscript)8) 8, log(subscript)16) 8, log(subscript)32) 8, log(subscript)64) 8…..
How do I find an expression for the nth term of the sequence in the form p/q, where p and q are integers.

Two things that could be helpful :
• $\displaystyle \log_b x = \frac{\ln x}{\ln b}$
• $\displaystyle \ln x^y = y\ln x$
Using this you can write each term of the sequence as a fraction. For example, for the fourth term :

$\displaystyle \log_{16} 8=\frac{\ln 8}{\ln 16}=\frac{\ln 2^3}{\ln 2^4}=\frac{3\ln 2}{4\ln 2}=\frac34$
• Oct 25th 2008, 07:40 AM
vale01
thanks
thank u so much!
that helps alot
• Oct 25th 2008, 10:01 AM
Soroban
Hello, vale01!

Quote:

Sequence: .$\displaystyle \log_2(8),\;\log_4(8),\; \log_8(8),\;\log_{16}(8),\;\log_{32}(8),\;\log_{64 }(8),\;\hdots$

Find an expression for the $\displaystyle n^{th}$ term in the form $\displaystyle \frac{p}{q}$, where $\displaystyle p$ and $\displaystyle q$ are integers.

Evaluate the logs ... Look for a pattern.

$\displaystyle \log_2(8) = x\quad\Rightarrow\quad 2^x = 8 \quad\Rightarrow\quad 2^x = 2^3 \quad\Rightarrow\quad x = 3$

$\displaystyle \log_4(8) = x \quad\Rightarrow\quad 4^x = 8\quad\Rightarrow\quad (2^2)^x = 2^3\quad\Rightarrow\quad 2x = 3 \quad\Rightarrow\quad x = \frac{3}{2}$

$\displaystyle \log_8(8) = x \quad\Rightarrow\quad 8^x = 8 \quad\Rightarrow\quad x = 1$

$\displaystyle \log_{16}(8) = x \quad\Rightarrow\quad 16^x = 8 \quad\Rightarrow\quad (2^4)^x = 2^3 \quad\Rightarrow\quad 4x = 3 \quad\Rightarrow\quad x = \frac{3}{4}$

$\displaystyle \log_{32}(8) = x \quad\Rightarrow\quad 32^x = 8 \quad\Rightarrow\quad (2^5)^x = 2^3 \quad\Rightarrow\quad 5x = 3 \quad\Rightarrow\quad x = \frac{3}{5}$

$\displaystyle \log_{64}(8) = x \quad\Rightarrow\quad 64^x = 8 \quad\Rightarrow\quad (2^6)^x = 2^3 \quad\Rightarrow\quad 6x = 3 \quad\Rightarrow\quad x = \frac{1}{2}$

$\displaystyle \begin{array}{c|ccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & \hdots \\ \hline \hline \\[-4mm] \text{We have:}& 3 & \frac{3}{2} & 1 & \frac{3}{4} & \frac{3}{5} & \frac{1}{2} & \hdots \\ \\[-4mm] \hline \\[-4mm] \text{That is:} & 3 & \frac{3}{2} & \frac{3}{3} & \frac{3}{4} & \frac{3}{5} & \frac{3}{6} & \hdots \\ \\[-4mm] \hline \end{array}$

And we see that the $\displaystyle n^{th}$ term is: .$\displaystyle a_n \:=\:\frac{3}{n}$

• Oct 25th 2008, 10:35 AM
vale01
thanks
thank u!