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.

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