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!
