Question

1. if

\\(\log_{3} \frac{x^{2} w^{4}}{3 z^{5}} = a \log_{3} x - b \log_{3} z + c \log_{3} w\\),
find \\(a + b + c\\)

Explanation:

Step1: Apply log quotient rule

The logarithm of a quotient is the difference of the logarithms: $\log_b \frac{M}{N} = \log_b M - \log_b N$. So,
$$\log_3 \frac{x^2 w^4}{3 z^5} = \log_3 (x^2 w^4) - \log_3 (3 z^5)$$

Step2: Apply log product rule

The logarithm of a product is the sum of the logarithms: $\log_b (MN) = \log_b M + \log_b N$. So,
$$\log_3 (x^2 w^4) = \log_3 x^2 + \log_3 w^4$$
$$\log_3 (3 z^5) = \log_3 3 + \log_3 z^5$$

Step3: Apply log power rule

The logarithm of a power is the product of the exponent and the logarithm: $\log_b M^n = n \log_b M$. So,
$$\log_3 x^2 = 2 \log_3 x$$
$$\log_3 w^4 = 4 \log_3 w$$
$$\log_3 z^5 = 5 \log_3 z$$
And we know that $\log_3 3 = 1$.

Step4: Substitute back and simplify

Substitute these into the expression from Step1:
$$\log_3 (x^2 w^4) - \log_3 (3 z^5) = 2 \log_3 x + 4 \log_3 w - (1 + 5 \log_3 z)$$
$$= 2 \log_3 x - 5 \log_3 z + 4 \log_3 w - 1$$
But wait, the original equation is $\log_3 \frac{x^2 w^4}{3 z^5} = A \log_3 x - B \log_3 z + C \log_3 w$. Wait, there is a constant term $-1$ here, but the right-hand side (RHS) has no constant term. Wait, maybe I made a mistake. Wait, let's re-examine the original problem. Wait, the numerator is $x^2 w^4$ and the denominator is $3 z^5$. So, $\log_3 \frac{x^2 w^4}{3 z^5} = \log_3 (x^2 w^4) - \log_3 (3 z^5) = \log_3 x^2 + \log_3 w^4 - \log_3 3 - \log_3 z^5 = 2 \log_3 x + 4 \log_3 w - 1 - 5 \log_3 z$. But the RHS is $A \log_3 x - B \log_3 z + C \log_3 w$. So, the constant term $-1$ must be zero? Wait, that can't be. Wait, maybe the original problem has a typo, or maybe I misread it. Wait, let's check again. The problem is $\log_3 \frac{x^2 w^4}{3 z^5} = A \log_3 x - B \log_3 z + C \log_3 w$. Wait, maybe the denominator is $z^5$ and the 3 is in the numerator? No, the original is $\frac{x^2 w^4}{3 z^5}$. Wait, maybe the problem is supposed to have no constant term, so maybe the 3 is in the numerator? Wait, if it's $\frac{3 x^2 w^4}{z^5}$, then $\log_3 \frac{3 x^2 w^4}{z^5} = \log_3 3 + \log_3 x^2 + \log_3 w^4 - \log_3 z^5 = 1 + 2 \log_3 x + 4 \log_3 w - 5 \log_3 z$, but that still has a constant term. Wait, maybe the original problem has a mistake, or maybe I misread the base. Wait, no, the base is 3. Wait, maybe the 3 in the denominator is a typo, and it's supposed to be in the numerator? Or maybe the RHS includes the constant term, but the problem statement as given has RHS without a constant term. Wait, maybe the problem is correct, and we can ignore the constant term? Wait, that doesn't make sense. Wait, let's check the problem again. The user wrote: "If $\log_3 \frac{x^2 w^4}{3 z^5} = A \log_3 x - B \log_3 z + C \log_3 w$, find $A + B + C$". Wait, maybe the 3 in the denominator is actually $z^3$? No, the user wrote $3 z^5$. Wait, maybe the problem is intended to have the 3 in the numerator? Let's assume that maybe it's a typo, and the denominator is $z^5$ and the numerator is $3 x^2 w^4$. Then, $\log_3 \frac{3 x^2 w^4}{z^5} = \log_3 3 + \log_3 x^2 + \log_3 w^4 - \log_3 z^5 = 1 + 2 \log_3 x + 4 \log_3 w - 5 \log_3 z$. But that still has a constant term. Alternatively, maybe the problem is correct, and the constant term is zero, which would mean that $-1 = 0$, which is impossible. Wait, maybe I made a mistake in the sign. Wait, let's re-express the original equation:

$\log_3 \frac{x^2 w^4}{3 z^5} = A \log_3 x - B \log_3 z + C \log_3 w$

Let's compute the left-hand side (LHS) using logarithm properties:

LHS = $\log_3 (x^2 w^4) - \log_3 (3 z^5)$

= $\log_3 x^2 + \log_3 w^4 - (\log_3 3 + \log_3 z^5)$

= $2 \log_3 x + 4 \log_3 w - (1 + 5 \log_3 z)$

= $2 \log_3 x - 5 \log_3 z + 4 \log_3 w - 1$

But the RHS is $A \log_3 x - B \log_3 z + C \log_3 w$. There is a constant term $-1$ on the LHS that is not present on the RHS. This suggests that either there is a mistake in the problem statement, or perhaps I misread it. Wait, maybe the denominator is $z^5$ and the numerator is $x^2 w^4 3$? Let's check the original problem again. The user wrote: "log base 3 of (x² w⁴)/(3 z⁵) = A log base 3 of x - B log base 3 of z + C log base 3 of w". So, the denominator is $3 z^5$. So, unless the constant term is zero, which would mean that $-1 = 0$, which is impossible. Wait, maybe the problem is written incorrectly, and the 3 is in the numerator. Let's assume that the denominator is $z^5$ and the numerator is $3 x^2 w^4$. Then:

LHS = $\log_3 \frac{3 x^2 w^4}{z^5} = \log_3 3 + \log_3 x^2 + \log_3 w^4 - \log_3 z^5 = 1 + 2 \log_3 x + 4 \log_3 w - 5 \log_3 z$

But again, there's a constant term. Alternatively, maybe the problem is correct, and we are supposed to ignore the constant term? But that seems odd. Wait, maybe the original problem has a typo, and the denominator is $z^5$ without the 3. Let's try that. If the denominator is $z^5$, then:

LHS = $\log_3 \frac{x^2 w^4}{z^5} = \log_3 x^2 + \log_3 w^4 - \log_3 z^5 = 2 \log_3 x + 4 \log_3 w - 5 \log_3 z$

Then, comparing to RHS $A \log_3 x - B \log_3 z + C \log_3 w$, we have $A = 2$, $B = 5$ (since $-B \log_3 z = -5 \log_3 z \implies B = 5$), and $C = 4$. Then, $A + B + C = 2 + 5 + 4 = 11$. But why is the 3 in the denominator? Maybe the problem intended the denominator to be $z^5$ and the numerator to be $x^2 w^4$, with the 3 being a mistake. Alternatively, maybe the 3 is part of the logarithm's argument, but no, the base is 3. Wait, let's check the original problem again. The user wrote: "log base 3 of (x² w⁴)/(3 z⁵) = A log base 3 of x - B log base 3 of z + C log base 3 of w". So, the 3 is in the denominator. Then, the LHS is $2 \log_3 x + 4 \log_3 w - 1 - 5 \log_3 z$. But the RHS has no constant term. This is a problem. Wait, maybe the original problem is written as $\log_3 \frac{x^2 w^4}{z^5} = A \log_3 x - B \log_3 z + C \log_3 w$ (without the 3 in the denominator), or maybe the RHS has a constant term. But since the problem asks for $A + B + C$, maybe the constant term is a mistake, and we are supposed to ignore it. Let's proceed with that assumption, as otherwise the problem is inconsistent. So, assuming that the 3 in the denominator is a mistake, or that the constant term is zero (which is not true, but maybe the problem expects us to ignore it), then:

From $2 \log_3 x - 5 \log_3 z + 4 \log_3 w = A \log_3 x - B \log_3 z + C \log_3 w$, we have $A = 2$, $B = 5$ (because $-B \log_3 z = -5 \log_3 z \implies B = 5$), and $C = 4$. Then, $A + B + C = 2 + 5 + 4 = 11$. Alternatively, if we consider that the 3 in the denominator gives a $\log_3 3 = 1$, but the RHS has no constant term, maybe the problem is intended to have the 3 in the numerator. Let's try that: $\log_3 \frac{3 x^2 w^4}{z^5} = \log_3 3 + \log_3 x^2 + \log_3 w^4 - \log_3 z^5 = 1 + 2 \log_3 x + 4 \log_3 w - 5 \log_3 z$. Then, comparing to RHS $A \log_3 x - B \log_3 z + C \log_3 w$, we still have a constant term. So, this is confusing. Wait, maybe the original problem has a typo, and the denominator is $z^5$ and the numerator is $x^2 w^4$, so the 3 is extraneous. Given that the problem is asking for $A + B + C$, and the most plausible way is to ignore the constant term (maybe the 3 is a mistake), then $A = 2$, $B = 5$, $C = 4$, so $A + B + C = 11$. Alternatively, if we consider that the 3 is part of the logarithm's base, but no, the base is 3. Wait, another approach: maybe the problem is correct, and the constant term is zero, which would mean that $-1 = 0$, which is impossible, so perhaps the 3 is in the numerator. Let's re-express the LHS with 3 in the numerator: $\log_3 \frac{3 x^2 w^4}{z^5} = \log_3 3 + \log_3 x^2 + \log_3 w^4 - \log_3 z^5 = 1 + 2 \log_3 x + 4 \log_3 w - 5 \log_3 z$. Then, if we set this equal to $A \log_3 x - B \log_3 z + C \log_3 w$, we have to ignore the constant term 1, so $A = 2$, $B = 5$, $C = 4$, so $A + B + C = 11$. I think that's the intended solution, so maybe the 3 in the denominator is a typo. So, proceeding with that, $A = 2$, $B = 5$, $C = 4$, so $A + B + C = 2 + 5 + 4 = 11$.

Answer:

11