# Logarithmic Functions

1. Sep 19, 2007

Last edited by a moderator: May 3, 2017
2. Sep 19, 2007

### Feldoh

First thing I would do is change the bases so they are equivalent.

3. Sep 19, 2007

How would you do that though =/

4. Sep 19, 2007

### rock.freak667

Change of base formula:

$$log_a b = \frac{log_c b}{log_c a}$$ where c is the new base

5. Sep 19, 2007

kk ty

6. Sep 19, 2007

The problem.

I'll write the problem out because your image is a little hard to see.

If $$\log_{2n}(1944)=\log_{n}(486\sqrt{2}),$$ determine the value of $$n^{3}$$

Let me know if you have more problems with this, im fresh on log functions.

Last edited: Sep 20, 2007
7. Sep 19, 2007

kk sooo log 1944/ log 2n = log 486/2 / log n

Do the logs cancel?

Last edited: Sep 19, 2007
8. Sep 19, 2007

Did you change to base 10? is that why your not placing the bases? and in the log to your right, I suppose thats not 486/2 but $$486\sqrt{2}$$?

Last edited: Sep 20, 2007
9. Sep 19, 2007

yea it is that but i didnt know how to do the thingie

10. Sep 19, 2007

do the logs cancel?

11. Sep 19, 2007

### Feldoh

Oh wow... That problem is interesting I actually think there is a better way to solve it.

12. Sep 20, 2007

### HallsofIvy

Since we don't know what you mean by "thingie", that tells us nothing!

You have already been told that you can write the equation as
$$\frac{log(1944)}{log(2n)}= \frac{log(468)}{log(n)}$$
Do you see that log(2n)= log(n)+ log(2)?
And that you can rewrite that equation as
$$\frac{1944}{log(n)+ log(2)}= \frac{log(468)}{log(n)}$$

I haven't specified the base of "log" because it could be anything. Choose base 10 if you like or natural logarithms. In any case, log(1944), log(2), and log(468) are numbers.

Could you solve an equation like
$$\frac{A}{x+ B}= \frac{C}{x}$$

and do you see why your equation is of that form?

13. Sep 20, 2007

Umm, by this don't you mean:
$$\frac{\log1944}{\log(n)+\log(2)}=\frac{log(486\sqrt{2})}{log(n)}$$

14. Sep 20, 2007

yea thanks but, i was able to find out what to do.

15. Sep 20, 2007

### HallsofIvy

Yes, don't know why I dropped the $\sqrt{2}$.