Question

question
evaluate:
\log_{9} \frac{1}{27}
answer
attempt 1 out of 2

Explanation:

Step1: Recall the change of base formula and logarithm properties. Let \( y = \log_{9}\frac{1}{27} \). By the definition of logarithms, this means \( 9^y=\frac{1}{27} \).

Step2: Express 9 and 27 as powers of 3. We know that \( 9 = 3^2 \) and \( 27 = 3^3 \), so \( \frac{1}{27}=3^{-3} \). Substituting these into the equation from Step 1, we get \( (3^2)^y = 3^{-3} \).

Step3: Simplify the left - hand side using the power of a power rule \((a^m)^n=a^{mn}\). So, \( (3^2)^y = 3^{2y} \). Now our equation is \( 3^{2y}=3^{-3} \).

Step4: Since the bases are the same and the exponential function \( f(x)=3^x \) is one - to - one, we can set the exponents equal to each other. So, \( 2y=-3 \).

Step5: Solve for y. Divide both sides of the equation \( 2y = - 3 \) by 2. We get \( y=-\frac{3}{2} \).

Answer:

\(-\frac{3}{2}\)