# Math Help - Logarithm bases

1. ## Logarithm bases

Im very confused with problem and i hope you can help me.

How to calculate the nth term of the fallowing sequences, giving the answer in the form p/q, where p, q ∈ Z

log_{1/5}(125) , log_{1/125}(125) , log_{1/625}(125)

log_{4}(64) , log_{8}(x) , log_{32}(64)

Also describe how to obtain the third answer in each row from the first two answers.

2. Originally Posted by erickvm123
Im very confused with problem and i hope you can help me.

How to calculate the nth term of the fallowing sequences, giving the answer in the form p/q, where p, q ∈ Z

log_{1/5}(125) , log_{1/125}(125) , log_{1/625}(125)
Each of these terms is a logarithm of 125 to bases of 1/5, $1/125= 1/5^3$, and $1/625= 1/5^4$. Frankly, I don't see a pattern unless there is a missing $log_{1/25}(125)$ in there. If that were the case, then the nth term would be $log_{1/5^n}(125)$.

The whole point of a logarithm is that $log_a(a^x)= x$. If the nth term is indeed $log_{1/5^n}(125)$, then $125= 5^3= (5^n)^{3/n}= (1/5^n)^{-3/n}$. The nth term would be $-\frac{3}{n}$.

log_{4}(64) , log_{8}(x) , log_{32}(64)

Also describe how to obtain the third answer in each row from the first two answers.
$log_8(x)$?? Surely, that is wrong. It must be $log_8(64)$ like the others. In that case, each is the logarithm of 64 to bases $4= 2^2$, $8= 2^3$, and $16= 2^4$. The nth term would be $log_{2^{n-1}}(64)$. $64= 8^2= (2^3)^2= 2^6= (2^{n-1})^{6/(n-1)}$. $log_{2^{n-1}}(64)= \frac{6}{n-1}$.

3. Hello, erickvm123!

I agree with HallsofIvy . . . the changing bases have no patten.
Besides the typos, I believe you left out some terms . . .

Calculate the $n^{th}$ term of the fallowing sequences,
giving the answer in the form $\tfrac{p}{q}$ where $p,q \in Z$

$\log_{4}(64),\quad \log_{8}(64), \quad \log_{16}(64),\quad \log_{32}(64),\quad \hdots$

Evaluating the logs, we have: . $3,\;2,\;\frac{3}{2},\;\frac{6}{5},\;\hdots$

Then we see the patten: . $\frac{6}{2},\;\frac{6}{3},\;\frac{6}{4},\;\frac{6} {5},\;\hdots$

Therefore, the $n^{th}$ term is: . $\frac{6}{n+1}$

4. well this is how it is in my paper that's why I'm confused, but thank you for finding the general term in p/q. And yeah it should be $log_{8} 64$ instead of $log_{8} (X)$