Verified answer

log_4 (x - 3) + log_4 (x - 3) = 1

2 log_4 (x - 3) = 1

log_4 (x - 3) = 1/2

4^(1/2) = x - 3

x - 3 = 2

x = 5

do 4 raised to both sides since the logs are base 4

4^(log4 (x - 3) + log4 (x - 3)) = 4^1

using the log property

4^(x + y) = 4^x 4^y

the equation becomes

4^(log4 (x - 3)) 4^(log4 (x - 3)) = 4

since 4^(log4(something)) = something

the equation becomes

(x - 3)(x - 3) = 4

x^2 - 6x + 9 = 4

or

x^2 - 6x + 5 = 0

solve using quadratic equation

x = 5 or x = 1

log4 (x - 3) + log4 (x - 3) = 1

domain (3, +∞]

2log4 (x - 3) = 1

log4 (x - 3)^2 = 1

(x - 3)^2 = 4^1

x^2 - 6x + 9 = 4

x^2 - 6x + 5 = 0

(x - 5)(x - 1) = 0

x = 5 (or x = 1, which is irrational for the original expression)

Graph of y = 2log4 (x - 3) is at:

http://s562.photobucket.com/albums/ss68/jintrater/...

log4 ((x- 3)²) = 1

(x - 3)² = 4

x = 5 and x = 1

However, only x = 5 works since you get a negative when you plug in 1 into x - 3.

properly in case you think of of the logarithms as numbers, then we are including 2 of the comparable variety jointly and that's the comparable as multiplying by using 2 so: log4(x-3) + log4(x-3) = 2*log4(x-3) = a million So this provides us: log4(x-3) = a million/2 Now if we flow lower back to the definition of the logarithm this fact is asserting that the skill to which the variety 4 could be raised to furnish (x-3) is a million/2, in different words: 4^(a million/2) = x-3 Now elevating a variety to the skill of a million/2 is the comparable as taking the sq. root, so: 2 = x-3 consequently: x = 5 desire this enables