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

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Answer 1 · 2016-02-24T15:16:15

#x=1#

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 #x > 0#.

#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 #log_3(x)# is one-to-one for #x in RR^+#, we take inverse of #log_3(x)# for both sides

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

#3x + 1 = 4x^2#

Solve the quadratic the usual way.

#4x^2 - 3x - 1 = 0#

#(4x + 1)(x - 1) = 0#

#x = 1 or x = -1/4#

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