# How do you solve #log_4 x^2 - log_4 (x+1) = 5#?

##### 1 Answer

#### Explanation:

**1) Determine the domain**

First of all, let's determin the domain of the two logarithmic expressions.

- As
#x^2 >= 0# for all#x in RR# ,#log_4(x^2)# is defined for all#x != 0# . #log_4(x+1)# is defined for#x + 1 > 0 <=> x > -1#

Thus, our domain is

**2) Solve the equation**

Now let's solve the equation. To start, use the logarithmic law

#log_a (m) - log_a(n) = log_a(m/n)#

Thus, you can transform your equation into:

#log_4(x^2 / (x+1)) = 5#

Now, the inverse function of

Thus, to "eliminate" the logarithmic term, you need to apply

#x^2/(x+1) = 4^5#

#x^2/(x+1) = 1024#

... multiply both sides with

#x^2 = 1024(x+1)#

... bring all the terms to the left side...

#x^2 - 1024x - 1024 = 0#

Solve the equation e.g. with the quadratic formula:

#x = (1024 +- sqrt(1024^2 + 4 * 1024))/2 ~~ (1024 +- 1025.998)/2#

#x ~~ - 0.999 " or " x ~~ 1024.999#

**3) Check the domain**

Now, we need to check if both

Indeed,