1. ## 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!!

2. Hi,
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 :
• $\log_b x = \frac{\ln x}{\ln b}$
• $\ln x^y = y\ln x$
Using this you can write each term of the sequence as a fraction. For example, for the fourth term :

$
\log_{16} 8=\frac{\ln 8}{\ln 16}=\frac{\ln 2^3}{\ln 2^4}=\frac{3\ln 2}{4\ln 2}=\frac34$

3. ## thanks

thank u so much!
that helps alot

4. Hello, vale01!

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

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

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

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

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

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

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

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

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

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

thank u!