How do you solve #log (2x + 1) - log (x - 2) = 1#?

1 Answer
Johnson Z.
Mar 16, 2016

#x=21/8#

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 left side of the equation.

#log(2x+1)-log(x-2)=1#

#log((2x+1)/(x-2))=1#

#2#. Use the log property, #log_color(purple)b(color(purple)b^color(orange)x)=color(orange)x#, to rewrite the right side of the equation.

#log((2x+1)/(x-2))=log(10)#

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

#(2x+1)/(x-2)=10#

#4#. Solve for #x#.

#2x+1=10(x-2)#

#2x+1=10x-20#

#8x=21#

#color(green)(|bar(ul(color(white)(a/a)x=21/8color(white)(a/a)|)))#

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