Answer
$x = 2$
$x = -\frac{1}{2}$
Work Step by Step
$$log_{2}x + log_{2}(2x - 3) = 1$$
Since both logarithmic terms have the same base, we can combine them using the Rules of Logarithms:
$log_{2}x + log_{2}(2x - 3) = log_{2}(x)(2x-3) = log_{2}(2x^{2} - 3x)$
We can now re-write the original equation as:
$$log_{2}(2x^{2} - 3x) = 1$$
And by changing the equation into exponential form:
$$2^{1} = (2x^{2} - 3x)$$
$0 = 2x^{2} - 3x - 2$; we can now use the quadratic formula $\frac{-b\frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$ to solve for $x$:
$$\frac{-(-3)\frac{+}{} \sqrt {(-3)^{2} - 4(2)(-2)}}{2(2)}$$
$$\frac{3\frac{+}{} \sqrt {9 + 16}}{4}$$
$$\frac{3\frac{+}{} \sqrt {25}}{4}$$
$$\frac{3\frac{+}{} 5}{4}$$
Where finally, we can say that $x = \frac{3+ 5}{4} = \frac{8}{4} = 2$ and $x = \frac{3- 5}{4} = \frac{-2}{4} = -\frac{1}{2}$
