Note that
[tex]log_{4} (x+3) = \frac{log_{2} (x+3)}{log_{2}4} [/tex]
Therefore, given
log₄(x+3) = log₂(2x), obtain
[tex] \frac{log_{2}(x+3)}{log_{2}4}=log_{2} (2x) \\ log_{2}(x+3) = log_{2}4 log_{2}(2x) \\ x+3 = (2x)^{log_{2}4} [/tex]
Because
log₂4 = log₂ 2² = 2 log₂ 2 = 2, therefore
x + 3 = (2x)² = 4x²
4x² - x - 3 = 0
This factorizes into
(4x + 3)(x - 1) = 0
Answer:
4x² - x - 3 = 0
or
4x + 3 = 0, and
x - 1 = 0
