How do you solve #log_3(x+6)-log_9x=log_9 2#?

1 Answer
Nov 2, 2016

Answer:

#x = O/#

Explanation:

Use the change of base formula, #log_a(n) = logn/loga#, to start the solving process.

#log(x + 6)/log3 - logx/log9 = log2/log9#

#log(x + 6)/log3 - logx/log3^2= log2/log3^2#

#log(x + 6)/log3 - logx/(2log3) = log2/(2log3)#

Put on a common denominator.

#(2log(x + 6))/(2log3) - logx/(2log3) = log2/(2log3)#

#log_9(x + 6)^2 - log_9x = log_9 2#

We now use the property #log_a(n) - log_a(m) = log_a(n/m)# to solve.

#log_9((x + 6)^2/x) = log_9 2#

#(x^2 + 12x + 36)/x = 2#

#x^2 + 12x + 36 = 2x#

#x^2 + 10x + 36 = 0#

#x = (-b +- sqrt(b^2 -4ac))/(2a)#

#x = (-10 +- sqrt(10^2 - (4 xx 1 xx 36)))/(2(1))#

#x = (-10 +- sqrt(-44))/2#

#x = (-10 +- 2sqrt(11)i)/2#

#x = -5 +- sqrt(11)i#

This does not satisfy the original equation, and hence there are no solutions.

Hopefully this helps!