Question

what is the value of $\log_{27}9$?
$\boldsymbol{\circ}$ $-\frac{3}{2}$
$\boldsymbol{\circ}$ $-\frac{2}{3}$
$\boldsymbol{\circ}$ $\frac{2}{3}$
$\boldsymbol{\circ}$ $\frac{3}{2}$

Explanation:

Step1: Recall the change of base formula or logarithmic property

We know that if we have $\log_{a}b$, we can also express it using the property of logarithms related to exponents. Let $y = \log_{27}9$. By the definition of logarithms, this means $27^{y}=9$.

Step2: Express both numbers as powers of 3

We know that $27 = 3^{3}$ and $9=3^{2}$. So we substitute these into the equation: $(3^{3})^{y}=3^{2}$.

Step3: Simplify the left - hand side using exponent rules

Using the exponent rule $(a^{m})^{n}=a^{mn}$, we get $3^{3y}=3^{2}$.

Step4: Set the exponents equal to each other

Since the bases are the same and the equation $3^{3y}=3^{2}$ holds, we can set the exponents equal: $3y = 2$.

Step5: Solve for y

Divide both sides of the equation $3y = 2$ by 3. We get $y=\frac{2}{3}$.

Answer:

$\frac{2}{3}$ (corresponding to the option $\frac{2}{3}$)