How do you solve #log_4 8 - log_4(x+6)=1#?

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Answer 1 · 2016-10-20T02:03:58

#log_color(magenta)4 color(red)8-log_color(magenta)4 color(blue)((x+6))=1#

Condense the log by using the log property #logcolor(red)a -logcolor(blue)b =log (color(red)a/color(blue)b)#

#log_color(magenta)4 (color(red)8/color(blue)(x+6))=1#

Next, turn the log into an exponential by the rule

#log_color(magenta)b color(orange)x=color(limegreen)a# becomes #color(magenta)b^color(limegreen)a= color(orange)x#.

#log_color(magenta)4 color(orange)((8/(x+6))color(red)=color(limegreen)1#

#color(magenta)4^color(limegreen)1=color(orange)(8/(x+6))#

#4=8/(x+6)#

#4(x+6)=8#

#4x+24=8#

#(4x)/4=-16/4#

#x=-4#