# How do you solve #log_9 (3x+1) = log_3 (x) + log_3 (2)#?

##### 1 Answer

Feb 24, 2016

#### Explanation:

Use the logarithmic identities

#log_a(b) = frac{log_x(b)}{log_x(a)}# for#x in RR^+# and

#log_x(ab) = log_x(a) + log_x(b)#

In this question, note that

#log_9(3x+1) = log_3(x) + log_3(2)#

#= log_3(3x+1)/log_3(9)#

#= log_3(3x+1)/2#

#log_3(3x+1) = 2log_3(x) + 2log_3(2)#

#= log_3(4x^2)#

Since

#3^(log_3(3x+1)) = 3^(log_3(4x^2))#

#3x + 1 = 4x^2#

Solve the quadratic the usual way.

Reject the negative answer as it is not in the domain.