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

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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!

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