How do you solve #log 3x=log 4+log (x-3)#?

1 Answer
Johnson Z.
Mar 17, 2016

#x=12#

Explanation:

#1#. Use the log property, #log_color(purple)b(color(red)m*color(blue)n)=log_color(purple)b(color(red)m)+log_color(purple)b(color(blue)n)# to simplify the right side of the equation.

#log(3x)=log4+log(x-3)#

#log(3x)=log(4(x-3))#

#log(3x)=log(4x-12)#

#2#. Since the equation now follows a "#log=log#" situation, where the bases are the same on both sides, rewrite the equation without the "log" portion.

#3x=4x-12#

#3#. Solve for #x#.

#3x-4x=-12#

#-x=-12#

#color(green)(|bar(ul(color(white)(a/a)x=12color(white)(a/a)|)))#

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