How do you solve the equation #log_a (4n)-2log_a x=log_ax# for n?

Question

1342 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-10T21:14:04

#n = 1/4x^3#

Start by using the rule #alogn = logn^a#.

#log_a(4n) - log_a(x^2) = log_a(x)#

Now, use the rule #logn - logm = log(n/m)#.

#log_a((4n)/x^2) = log_a(x)#

We know that if #loga = logb#, then #a = b#, so:

#(4n)/x^2 = x#

#4n = x^3#

#n = 1/4x^3#

Hopefully this helps!