Question

evaluating logarithms
evaluate: \\(\log_{27} 9\\)
\\(-\frac{1}{3}\\) \\(\frac{3}{2}\\) \\(\frac{2}{3}\\)
\\(\frac{1}{3}\\)

Explanation:

Step1: Express as exponential equation

Let $\log_{27} 9 = x$, so $27^x = 9$

Step2: Rewrite to same base

$27=3^3$, $9=3^2$, so $(3^3)^x = 3^2$

Step3: Simplify exponents

$3^{3x} = 3^2$, set exponents equal: $3x=2$

Step4: Solve for x

$x=\frac{2}{3}$

Answer:

$\frac{2}{3}$