How do you solve #log_[5] (24x + 4) - 2log_[5] (x - 2) = 2#?

1 Answer
Apr 2, 2016

Answer:

#x=4#

Explanation:

#1#. Use the logarithmic property, #log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m)#, to rewrite #2log_5(x-2)#.

#log_5(24x+4)-2log_5(x-2)=2#

#log_5(24x+4)-log_5((x-2)^2)=2#

#2#. Use the logarithmic 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_5((24x+4)/(x-2)^2)=2#

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

#log_5((24x+4)/(x-2)^2)=log_5(5^2)#

#4#. 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.

#(24x+4)/(x-2)^2=5^2#

#5#. Isolate for #x#.

#24x+4=25(x-2)^2#

#24x+4=25(x^2-4x+4)#

#24x+4=25x^2-100x+100#

#color(darkorange)25x^2# #color(turquoise)(-124)x# #color(violet)(+96)=0#

#6#. Use the quadratic formula to solve for #x#.

#color(darkorange)(a=25)color(white)(XXXXX)color(turquoise)(b=-124)color(white)(XXXXX)color(violet)(c=96)#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(color(turquoise)(-124))+-sqrt((color(turquoise)(-124))^2-4(color(darkorange)(25))(color(violet)(96))))/(2(color(darkorange)(25)))#

#x=(124+-sqrt(15376-9600))/50#

#x=(124+-sqrt(5776))/50#

#x=(124+-76)/50#

#x=(124+76)/50color(white)(i),color(white)(i)(124-76)/50#

#x=200/50color(white)(i),color(white)(i)48/50#

#color(green)(|bar(ul(color(white)(a/a)x=4color(white)(i),color(white)(i)color(red)cancelcolor(green)(24/25)color(white)(a/a)|)))#

However, if you substitute #x=24/25# back into the original equation, you will find that you will end up taking the logarithm of a #color(red)("negative number")#, which is #color(red)("not possible")#.

For example:

#log_5(24x+4)-2log_5(x-2)=2#

#log_5(24(24/25)+4)-2log_5(24/25-2)=2#

#log_5(676/25)-2log_5(color(red)(-26/25))=2#

For this reason, the correct and only solution to the given logarithmic equation is #x=4#.