1. log 3 + log (5x - 2) = log (9x + 6)

2. 2 log (2x - 1) = 2 log (3x + 1) - log 4

3. log_2 (x - 4) + log_2 (x - 3) = 1

2. Originally Posted by jwhite09
1. log 3 + log (5x - 2) = log (9x + 6)
Look at the left hand side:

log 3 + log (5x - 2) = log(3(5x-2)) = log(15x-6),

so:

15x-6 = 9x+6.

Now try the same sort of thing on the remainder.

RonL

3. Originally Posted by jwhite09

1. log 3 + log (5x - 2) = log (9x + 6)
$\displaystyle Log (3 \times (5x - 2)) = Log (9x + 6)$

$\displaystyle 15x - 6 = 9x + 6$

$\displaystyle 6x = 12$

$\displaystyle x = 2$

Originally Posted by jwhite09

2. 2 log (2x - 1) = 2 log (3x + 1) - log 4
$\displaystyle log (2x-1)^2 = log (3x + 1)^2 - log (4)$

$\displaystyle log (2x-1)^2 = log \frac{(3x + 1)^2}{4}$

$\displaystyle (2x - 1)^2 = \frac{(3x + 1)^2}{4}$

$\displaystyle 4(2x - 1)^2 = (3x + 1)^2$

$\displaystyle 4(4x^2 - 4x + 1) = 9x^2 + 6x + 1$

$\displaystyle 16x^2 - 16x + 4 = 9x^2 + 6x + 1$

$\displaystyle 7x^2 - 22x + 3 = 0$

You can solve for x here, I'm lazy.

Originally Posted by jwhite09

3. log_2 (x - 4) + log_2 (x - 3) = 1
$\displaystyle log_2 ((x - 4)(x - 3)) = 1$

$\displaystyle log_2 (x^2 - 7x + 12) = 1$

$\displaystyle 2^1 = x^2 - 7x + 12$

$\displaystyle 0 = x^2 - 7x + 10$

$\displaystyle 0 = (x - 5)(x - 2)$

$\displaystyle x = 5 \ or \ x = 2$