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
Please Help!
$\displaystyle Log (3 \times (5x - 2)) = Log (9x + 6)$
$\displaystyle 15x - 6 = 9x + 6$
$\displaystyle 6x = 12$
$\displaystyle x = 2$
$\displaystyle log (2x-1)^2 = log (3x + 1)^2 - log (4)$
$\displaystyle log (2x-1)^2 = log \frac{(3x + 1)^2}{4}$
$\displaystyle (2x - 1)^2 = \frac{(3x + 1)^2}{4}$
$\displaystyle 4(2x - 1)^2 = (3x + 1)^2$
$\displaystyle 4(4x^2 - 4x + 1) = 9x^2 + 6x + 1$
$\displaystyle 16x^2 - 16x + 4 = 9x^2 + 6x + 1$
$\displaystyle 7x^2 - 22x + 3 = 0$
You can solve for x here, I'm lazy.
$\displaystyle log_2 ((x - 4)(x - 3)) = 1$
$\displaystyle log_2 (x^2 - 7x + 12) = 1$
$\displaystyle 2^1 = x^2 - 7x + 12$
$\displaystyle 0 = x^2 - 7x + 10$
$\displaystyle 0 = (x - 5)(x - 2)$
$\displaystyle x = 5 \ or \ x = 2$