Answer
$x=1$
Work Step by Step
Recall the power property of logarithms (pg. 462):
$\log_b{m^n}=n\log_b{m}$
Applying this property, we get:
$\log_2 4^{3/2}-\log_2 x^{1/2}=3$
$\log_2 (4^{1/2})^3-\log_2 x^{1/2}=3$
$\log_2 (2^3)-\log_2 \sqrt{x}=3$
$\log_2 (8)-\log_2 \sqrt{x}=3$
Next, recall the quotient property of logarithms (pg. 462):
$\log_b{\frac{m}{n}}=\log_b{m}-\log_b{n}$
Applying this property to our last equation, we get:
$\log_2 \frac{8}{\sqrt{x}}=3$
Recall the definition of a logarithm (pg. 451):
$\log_{b}{x}=y$ iff $b^y=x$
Applying this definition, we get:
$2^3=\frac{8}{\sqrt{x}}$
$8\cdot \sqrt{x}=8$
$\sqrt{x}=1$
$x=1^2$
$x=1$
We check the answer:
$\frac{3}{2}\log_2 4-\frac{1}{2}\log_2 1=3$
$\frac{3}{2}*2-\frac{1}{2}*0=3$
$3-0=3$
$3=3$
