algebra 2 help?

Question

Solve: log4(x^2) - log4(4x - 4) = 0

Answer 1

As a rule:

Log.x(y^h) = h*Log.x(y)

So now


Log.4(x^2) = 2*Log.4(x)

Therefore:

2*Log.4(x) - Log.4(4x - 4) = 0

Another logarithmic rule is:

Log.x(y) - Log.x(z) = Log.x(y/z)

So now:

2*Log.4(x) - Log.4(4x - 4) = 0

2 * (Log.4(x / (4x-4) ) ) = 0
2 * (Log.4(1 / 4-4 ) ) = 0
2 * (Log.4(1 / 0 ) ) = 0
2 * (Log.4(1) ) = 0
2 * Log.4(1) = 0
Log.4(1^2) = 0
Log.4(1) = 0
4^0 = 1 √

Therefore x = 1


Good luck.

Answer 2

log4(x^2) - log4(4x - 4) = 0
log4 [(x^2)/(4x - 4)] = 0
4^0 = [(x^2)/(4x - 4)]
4x - 4 = x^2
x^2 - 4x + 4 = 0
(x-2)^2 = 0
x = 2

Answer 3

In the following,log is taken to mean log to base 4:-
log (x²) - log (4x - 4) = 0
log [ x² / (4x - 4) ] = 0
x² / (4x - 4) = 4^0
x² / (4x - 4) = 1
x² = 4x - 4
x² - 4x + 4 = 0
(x - 2)² = 0
x = 2

Answer 4

log_4 (x^2) - log_4 (4x - 4) = 0

add log_4 (4x - 4) for both sides

log_4 (x^2) = log_4 (4x - 4)

if log a = log b, then a = b
x^2 = 4x - 4

x^2 - 4x + 4 = 0

(x - 2)^2 = 0

x = 2

Answer 5

4? i think thats it I only did it on paper.