How do you solve #log_3x=log_9 7x-6#?

1 Answer
Mar 21, 2016

#x=7/531441#

Explanation:

#1#. Start by moving all logs to the left side of the equation.

#log_3(x)=log_9(7x)-6#

#log_3(x)-log_9(7x)=-6#

#2#. Use the change of base formula, #log_color(blue)n(color(red)m)=(log_color(purple)b(color(red)m))/(log_color(purple)b(color(blue)n))#, to rewrite #log_9(7x)# with a base of #3#.

#log_3(x)-(log_3(7x))/(log_3(9))=-6#

#3#. Use the log property, #log_color(purple)b(color(purple)b^color(orange)x)=color(orange)x#, to rewrite #log_3(9)#.

#log_3(x)-(log_3(7x))/(log_3(3^2))=-6#

#log_3(x)-(log_3(7x))/2=-6#

#4#. Use the log property, #log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m)#, to rewrite #(log_3(7x))/2#.

#log_3(x)-1/2(log_3(7x))=-6#

#log_3(x)-log_3((7x)^(1/2))=-6#

#5#. 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_3(x/((7x)^(1/2)))=-6#

#log_3(x^(1/2)/7^(1/2))=-6#

#6#. 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_3(x^(1/2)/7^(1/2))=-log_3(3^6)#

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

#log_3(x^(1/2)/7^(1/2))=log_3(3^-6)#

#8#. 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.

#x^(1/2)/7^(1/2)=3^-6#

#9#. Solve for #x#.

#x^(1/2)/7^(1/2)=1/3^6#

#x^(1/2)/7^(1/2)=1/729#

#x^(1/2)=7^(1/2)/729#

#(x^(1/2))^2=(7^(1/2)/729)^2#

#color(green)(|bar(ul(color(white)(a/a)x=7/531441color(white)(a/a)|)))#