# Logarithmic Functions

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

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

How would you do that though =/

rock.freak667
Homework Helper
Change of base formula:

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

kk ty

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.

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

Do the logs cancel?

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

Do the logs cancel?

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 $$486\sqrt{2}$$?

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

do the logs cancel?

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

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

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?

Hallsofivy said:
$$\frac{1944}{log(n)+ log(2)}= \frac{log(468)}{log(n)}$$

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

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

HallsofIvy
$$\frac{\log1944}{\log(n)+\log(2)}=\frac{log(486\sqrt{2})}{log(n)}$$
Yes, don't know why I dropped the $\sqrt{2}$.