Write the log expression as a single log.
${\log }_2\frac{1}{9x}+2{\log }_{2}3x$

• A. ${\log }_{2}x$
• B. ${\log }_2\frac{1}{3x}$
• C. ${\log }_{2}1$
• D. ${\log }_{2}3x$

Answer: (A)

Verified step by step guidance

1. Start by using the property of logarithms that allows you to move the coefficient in front of a logarithm as an exponent inside the logarithm: \( a \cdot \log_b(c) = \log_b(c^a) \). Apply this to \( 2\log_2(3x) \) to get \( \log_2((3x)^2) \).
2. Next, simplify \( (3x)^2 \) to \( 9x^2 \). So, \( 2\log_2(3x) \) becomes \( \log_2(9x^2) \).
3. Now, combine the logarithms using the property \( \log_b(a) + \log_b(c) = \log_b(a \cdot c) \). Apply this to \( \log_2\left(\frac{1}{9x}\right) + \log_2(9x^2) \).
4. Multiply the expressions inside the logarithms: \( \frac{1}{9x} \cdot 9x^2 = x \).
5. Finally, write the expression as a single logarithm: \( \log_2(x) \).