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

1 Answer
Feb 8, 2016

Answer:

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

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 #x > -1# and #x != 0#.

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 #log_4(x)# is #4^x# which means that both #log_4(4^x) = x# and #4^(log_4(x)) = x# hold.

Thus, to "eliminate" the logarithmic term, you need to apply #4^x# to both sides of the equation:

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

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

... multiply both sides with #(x+1)#...

#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 #x# fit into our domain.

Indeed, #x > -1# and #x != 0# holds for both of them, so both are solutions.