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

1 Answer
Nov 10, 2016

n = 1/4x^3

Explanation:

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!