Find all real solutions.

Question

Find all real solutions to the equation log4(3x^2) - log4(4x+8) = -1

Be sure to check any potential solutions in the original equation.

Answer 1

log₄(3x²) - log₄(4x + 8) = -1

log₄ [3x² / (4x+8)] = -1

3x² / (4x+8) = 4^-1 = 1/4

3x² / (4x+8) = 1/4

(3x²)(4) = (4x+8)(1)

12x² - 4x - 8 = 0

3x² - x - 2 = 0

(3x + 2)(x - 1) = 0 So, x = -2/3, 1

Answer 2

log base 4 of (3x^2) - log base 4 of (4x+8) = -1

combine logs

log base 4 of (3x^2/4x+1) =-1

rewrite in exponential form

4^-1 = 3x^2/(4x+8) = 1/4

cross multiply

12x^2 = 4x+8

divide by 4

3x^2 = x +2

3x^2 -x -2 = 0

factor, complete the square, use the quadratic formula or graph the equation

(3x+2)(x-1) = 0

set each factor = 0 and solve for x

x = -2/3, 1

check the solutions. plug them into the original equations and see if they work

log base 4 (3) - log base 4 (4+8) = log base 4 (3/12) = log base 4 (1/4)

4^-1 = 1/4. it works

try -2/3

log base 4 (3(4/9) - log base 4 (-8/3+8) = log base 4 ((12/9)/(16/3)) = log base 4 (1/4)

it also works

Answer 3

First contract the left side using Laws of Logrithms.

Then convert to exponential form

Set eqution to zero and solve using Zero Product Rule.