Question

what is the value of \\(\log_{27} 9\\)?

Explanation:

Step1: Recall the change of base formula

The change of base formula for logarithms is $\log_{a}b=\frac{\log_{c}b}{\log_{c}a}$ (where $c>0,c
eq1$). We can also use the property of logarithms that if $a^{x}=b$, then $\log_{a}b = x$. Let's express 27 and 9 as powers of 3. We know that $27 = 3^{3}$ and $9=3^{2}$.

Step2: Apply the logarithm property

Let $x=\log_{27}9$. By the definition of logarithms, this means $27^{x}=9$. Substituting $27 = 3^{3}$ and $9 = 3^{2}$, we get $(3^{3})^{x}=3^{2}$. Using the exponent rule $(a^{m})^{n}=a^{mn}$, we have $3^{3x}=3^{2}$. Since the bases are the same, we can set the exponents equal: $3x = 2$.

Step3: Solve for x

To solve for $x$, we divide both sides of the equation $3x=2$ by 3. So $x=\frac{2}{3}$.

Answer:

$\frac{2}{3}$ (assuming the last option is $\frac{2}{3}$, the answer is the option with $\frac{2}{3}$)