How do you solve # log2 x = log (x + 16) - 1#?

1 Answer
Mar 16, 2016

#x=16/19#

Explanation:

#1#. Start by bringing #log(x+16)# to the left side of the equation.

#log(2x)=log(x+16)-1#

#log(2x)-log(x+16)=-1#

#2#. Using 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)#, simplify the equation.

#log((2x)/(x+16))=-1#

#3#. Convert to exponential form.

#10^(-1)=((2x)/(x+16))#

#4# Solve for #x#.

#1/10=((2x)/(x+16))#

#(x+16)/10=2x#

#x+16=20x#

#19x=16#

#color(green)(|bar(ul(color(white)(a/a)x=16/19color(white)(a/a)|)))#