How do you solve #log_4x- log_4(x-1)= 1/2#?

1 Answer
Johnson Z.
Feb 27, 2016

#x=2#

Explanation:

#1#. Recall the log quotient rule: #log_color(red)b(color(blue)m/color(purple)n)=log_color(red)b(color(blue)m)-log_color(red)b(color(purple)n)#. Using the formula, rewrite the left side of the equation.

#log_4(x)-log_4(x-1)=1/2#

#log_4(x/(x-1))=1/2#

#2#. Rewrite the equation in exponential form.

#4^(1/2)=x/(x-1)#

#3#. Solve for #x#.

#2=x/(x-1)#

#x=2(x-1)#

#x=2x-2#

#color(green)(x=2)#

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