How do you evaluate #\log _ { 2} ( 4- 2b ) = \log _ { 2} ( - 3b - 3)#?

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Answer 1 · 2017-08-09T16:23:45

#b = - 7#

We have: #log_(2)(4 - 2 b) = log_(2)(- 3 b - 3)#

#Rightarrow log_(2)(4 - 2 b) - log_(2)(- 3 b - 3) = 0#

Using the laws of logarithms:

#Rightarrow log_(2)(frac(4 - 2 b)(- 3 b - 3)) = 0#

#Rightarrow frac(4 - 2 b)(- 3 b - 3) = 2^(0)#

#Rightarrow frac(4 - 2 b)(- 3 b - 3) = 1#

#Rightarrow 4 - 2 b = - 3 b - 3#

#Rightarrow 3 b - 2 b = - 4 - 3#

#therefore b = - 7#

Therefore, the solution to the equation is #b = - 7#.

Answer 2 · 2017-08-09T16:28:56

#b=-7#

#"using the "color(blue)"law of ogarithms"#

#•color(white)(x)log_bx=log_byrArrx=y#

#rArr4-2b=-3b-3#

#rArrb=-7#