How do you solve #log_9 4+2 log_9 5=log_9 w#?

1 Answer
Johnson Z.
Mar 21, 2016

#w=100#

Explanation:

#1#. 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 #2log_(9)5#.

#log_(9)4+2log_(9)5=log_(9)w#

#log_(9)4+log_(9)5^2=log_(9)w#

#2#. 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_(9)(4*5^2)=log_(9)w#

#log_(9)(100)=log_(9)w#

#3#. 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.

#100=w#

#color(green)(|bar(ul(color(white)(a/a)w=100color(white)(a/a)|)))#

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