How do you solve for y in #Ln(y - 7) - ln(7) = x + ln(x)#?

1 Answer
Mar 28, 2016

#y=y=e^(x)7x+7#

Explanation:

#1#. Start by using the natural logarithmic property, #ln_color(purple)b(color(red)m/color(blue)n)=ln_color(purple)b(color(red)m)-ln_color(purple)b(color(blue)n)# to simplify the left side of the equation.

#ln(y-7)-ln(7)=x+ln(x)#

#ln((y-7)/7)=x+ln(x)#

#2#. Use the natural logarithmic property, #ln_color(purple)b(color(purple)b^color(orange)x)=color(orange)x#, to rewrite #x# on the right side of the equation.

#ln((y-7)/7)=ln(e^x)+ln(x)#

#3#. Use the natural logarithmic property, #ln_color(purple)b(color(red)m*color(blue)n)=ln_color(purple)b(color(red)m)+ln_color(purple)b(color(blue)n)# to simplify the right side of the equation.

#ln((y-7)/7)=ln(e^x##x)#

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

#(y-7)/7=e^(x)x#

#5#. Solve for #y#.

#y-7=e^(x)7x#

#color(green)(|bar(ul(color(white)(a/a)y=e^(x)7x+7color(white)(a/a)|)))#