fill in the table
x log base A of X
2
3 n
4
5
6
7 1
8 m
9
10 p
98
i know that a = 7
i think x =2 is m/3 ...im not sure tho
Well, for x = 8 we can get
$\displaystyle log_7(8) = log_7(2^3) = 3 \cdot log_7(2) = m$
So
$\displaystyle log_7(2) = \frac{m}{3}$
You were right.
We can pull the same thing with x = 4 and x = 9.
Once you have that you can do $\displaystyle log_7(6) = log_7(2 \cdot 3) = log_7(2) + log_7(3)$
And then do something similar for x = 5, because we know that $\displaystyle 10 = 2 \cdot 5$.
For x = 98, break 98 down into its prime factors. Then you'll see what to do.
-Dan