Answer to Question #130268 in Algebra for jem belly

Question

Question 2: Solve the following equation:

6 + log ⁡ ( 16 x 2 − 64 ) = log ⁡ ( 4 x + 8 ) + 8

A . x = 5.456

B . x = 3.859

C . x = 3.847

D . x = 27
E. none of the above
Choose the correct answer (A - E):

Answer 1

6 + log ⁡ (16x² − 64) = log ⁡ (4x + 8) + 8

Move terms

$log ⁡ (16x^2−64) - log (4x + 8) = 8 - 6$

$log ⁡ (16x^2 − 64) - log (4x + 8) = 2$

From

$log (x) - log (y) = log (x/y)$

$log ⁡ [(16x^2 − 64)/(4x + 8)] = 2$

Factor 4 out from expression.

$log ⁡ [4(4x^2 − 16)/4(x + 2)] = 2$

$log ⁡ [4x4(x^2 − 4)/4(x + 2)] = 2$

$log ⁡ [4(x^2 − 4)/(x + 2)] = 2$

Since from: $a^2 - b^2 = (a-b) (a+b)$

$log ⁡ [4(x − 2)(x+2)/4(x + 2)] = 2$

The expression simplifies to:

$log ⁡ [4(x-2)] = 2$

Since 2 = log 100

$log ⁡ [4(x-2)] = log100$

$[4(x-2)] = 100$

$4x-8= 100$

$4x= 108$

$x=27$

The answer is D = 27