# Thread: Logs and more logs....

1. ## Logs and more logs....

I have to find an expression for the nth term for all these sequences and write the expressions in the form $\frac {p}{q}$ where $p$, $q$, exists on integers.

Here are the sequences:
$\log_2{8}$, $\log_4{8}$, $\log_8{8}$, $\log_{16}{8}$, $\log_{32}{8}$, $\log_{64}{8}$, $\log_{128}{8}$

$\log_{m}{m^k}$, $\log_{m^2}{m^k}$, $\log_{m^3}{m^k}$, $\log_{m^4}{m^k}$, $\log_{m^5}{m^k}$, $\log_{m^6}{m^k}$, $\log_{m^7}{m^k}$

I don't know where to begin or how to approach this. Help?

2. Originally Posted by tinytiger
I have to find an expression for the nth term for all these sequences and write the expressions in the form $\frac {p}{q}$ where $p$, $q$, exists on integers.

Here are the sequences:
$\log_2{8}$, $\log_4{8}$, $\log_8{8}$, $\log_{16}{8}$, $\log_{32}{8}$, $\log_{64}{8}$, $\log_{128}{8}$

$\log_{m}{m^k}$, $\log_{m^2}{m^k}$, $\log_{m^3}{m^k}$, $\log_{m^4}{m^k}$, $\log_{m^5}{m^k}$, $\log_{m^6}{m^k}$, $\log_{m^7}{m^k}$

I don't know where to begin or how to approach this. Help?
I'll take the first example:

1. You are supposed to know that $8 = 2^3$

2. Use the base-change-formula of logarithms:

$\log_2(8) = \dfrac{\ln(8)}{\ln(2)} = \dfrac{3 \ln(2)}{\ln(2)} = 3$

$\log_4(8) = \dfrac{\ln(8)}{\ln(4)} = \dfrac{3 \ln(2)}{ 2\ln(2)} = \dfrac32$

$\log_8(8) = \dfrac{\ln(8)}{\ln(8)} = \dfrac{3 \ln(2)}{ 3\ln(2)} = \dfrac33 = 1$

$\log_{16}(8) = \dfrac{\ln(8)}{\ln(16)} = \dfrac{3 \ln(2)}{ 4\ln(2)} = \dfrac34$

Do you see the pattern?

The second example should be done in just the same way.

3. So for the first sequence the expression would be:

$\log_{2^n}(8) = \dfrac {\ln{2^3}}{\ln{2^n}} = \dfrac{3 \ln(2)}{n \ln(2)} = \dfrac {3}{n}$

And the last sequence the expression would be:

$\log_{m^n}{m^k} = \dfrac {\ln{m^k}}{\ln{m^n}} = \dfrac{k \ln(m)}{n \ln(m)} = \dfrac {k}{n}$

It makes sense now.

4. Originally Posted by tinytiger
So for the first sequence the expression would be:

$\log_{2^n}(8) = \dfrac {\ln{2^3}}{\ln{2^n}} = \dfrac{3 \ln(2)}{n \ln(2)} = \dfrac {3}{n}$

And the last sequence the expression would be:

$\log_{m^n}{m^k} = \dfrac {\ln{m^k}}{\ln{m^n}} = \dfrac{k \ln(m)}{n \ln(m)} = \dfrac {k}{n}$

It makes sense now.
Correct!

5. I just have one last question.

I have to obtain the third answer from the first two answers.

$\log_4{64}$ and $\log_8{64}$ must some how be calculated to get $\log_{32}{64}$.

I know that $\log_4{64} = 3$ and $\log_8{64} = 2$ and $\log_{32}{64} = 1.2$. I also recognize that the bases in the first 2 logarithms, 4 which is $2^2$, and 8, which is $2^3$, must have a connection with the last one, 32, which is $2^5$. I can't seem to figure it out though.

6. Originally Posted by tinytiger
I just have one last question.

I have to obtain the third answer from the first two answers.

$\log_4{64}$ and $\log_8{64}$ must some how be calculated to get $\log_{32}{64}$.

I know that $\log_4{64} = 3$ and $\log_8{64} = 2$ and $\log_{32}{64} = 1.2$. I also recognize that the bases in the first 2 logarithms, 4 which is $2^2$, and 8, which is $2^3$, must have a connection with the last one, 32, which is $2^5$. I can't seem to figure it out though.
You probably have seen that $\ln(4)+\ln(8)=2\ln(2)+3\ln(2)=5\ln(2)=\ln(32)$

Using the same method as in the previous posts you get:

$\log_{32}(64)=\dfrac{6\ln(2)}{5\ln(2)}=\dfrac{6\ln (2)}{2\ln(2)+3\ln(2)}= \dfrac65$

7. I, personally, wouldn't have used "ln", or the change of base formula for thes problems.

Instead use the basic definition of $log_a(x)$ as the inverse to $a^x$, and vice-versa. That means that $log_a(a^x)= x$.

So $log_2(8)= log_2(2^3)= 3$ immediately.

And, in general, for any n, $2^3= (2^3)^{n(1/n)}= 2^{3n(1/n)}= (2^n)^{3/n}$ and so $log_{2^n}(2^3)= log_{2^n}((2^n)^{3/n})= \frac{3}{n}$.

$m^3= (m^k)^{n(1/n)}= m^{nk/n}= (m^n)^{k/n}$ so $log_{m^n}(m^k)= log_{m^n}((m^n)^{k/n})= k/n$.

8. Thanks for the help!