Question

log\frac{9x^{7}y}{x^{2}y^{4}} = a\log3 + b\log x + c\log y

Explanation:

Step1: Apply log quotient rule

The logarithm of a quotient is the difference of the logarithms: $\log\frac{a}{b}=\log a - \log b$. So, $\log\frac{9x^{7}y}{x^{2}y^{4}}=\log(9x^{7}y)-\log(x^{2}y^{4})$.

Step2: Apply log product rule

The logarithm of a product is the sum of the logarithms: $\log(ab)=\log a+\log b$. So, $\log(9x^{7}y)=\log 9+\log x^{7}+\log y$ and $\log(x^{2}y^{4})=\log x^{2}+\log y^{4}$.

Step3: Simplify $\log 9$

Since $9 = 3^{2}$, $\log 9=\log 3^{2}$.

Step4: Apply log power rule

The logarithm of a power is the product of the exponent and the logarithm: $\log a^{n}=n\log a$. So, $\log 3^{2}=2\log 3$, $\log x^{7}=7\log x$, $\log x^{2}=2\log x$, and $\log y^{4}=4\log y$.

Step5: Substitute back and simplify

Substitute the simplified forms back into the equation:

$$
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$$

Comparing with $A\log 3 + B\log x + C\log y$, we get $A = 2$, $B = 5$, $C=-3$.

Answer:

$A = 2$, $B = 5$, $C=-3$