Solve: #2=log_6x + log_6 (x+9)# ?

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Answer 1 · 2017-01-22T02:42:30

#x=3#

#2=log_6x + log_6 (x+9)#

Remember #log_6 6= 1#

#:.2log_6 6 = log_6 x + log_6 (x+9)#

Also remember that #log_a b + log_a c = log_a (bxxc)#

#:. 2log_6 6 = log_6 (x(x+9))#

#log_6 6^2 = log_6(x^2+9x)#

#:. 36= x^2+9x#

#x^2+9x-36 =0#

#(x+12)(x-3)=0 -> x=-12 or x=3#

But #log_6 x# is only defined for #x>0#

#:. x=3#