Question

determine the value of the exponent needed to get the final answer.
1
$(6xy^{4})^{\square} = 36x^{2}y^{8}$
$\square = \underline{\quad\quad\quad}$
2
$2(-3n^{\square}p \bullet np)^{4} = 162n^{24}p^{8}$
$-3n^{\square} = \underline{\quad\quad\quad}$
3
$(6x^{\square}y^{\square})^{3} = 216x^{15}y^{9}$
$x^{\square} = \underline{\quad\quad\quad}$
$y^{\square} = \underline{\quad\quad\quad}$

Explanation:

Problem 1

Step1: Apply power of a product rule

Let the exponent be \( n \). Then \( (6xy^4)^n = 6^n x^n y^{4n} \).

Step2: Compare coefficients and exponents

We know \( 6^n = 36 \), \( x^n = x^2 \), \( y^{4n} = y^8 \).
For \( 6^n = 36 \), since \( 6^2 = 36 \), \( n = 2 \).
Check \( x^n = x^2 \): \( n = 2 \) works.
Check \( y^{4n} = y^8 \): \( 4n = 8 \Rightarrow n = 2 \).

Step1: Simplify inside the parentheses

First, simplify \( -3n^{\square}p \cdot np = -3n^{\square + 1}p^2 \).

Step2: Apply power of a product rule

Then \( 2(-3n^{\square + 1}p^2)^4 = 2 \cdot (-3)^4 n^{4(\square + 1)} p^{8} \).

Step3: Calculate coefficients and exponents

\( (-3)^4 = 81 \), so \( 2 \cdot 81 = 162 \), which matches the coefficient.
For the exponent of \( n \): \( 4(\square + 1) = 24 \).
Solve \( 4(\square + 1) = 24 \): \( \square + 1 = 6 \Rightarrow \square = 5 \).

Step1: Apply power of a product rule

Let the exponent of \( x \) be \( m \) and of \( y \) be \( k \). Then \( (6x^m y^k)^3 = 6^3 x^{3m} y^{3k} \).

Step2: Compare coefficients and exponents

\( 6^3 = 216 \), which matches.
For \( x \): \( 3m = 15 \Rightarrow m = 5 \).
For \( y \): \( 3k = 9 \Rightarrow k = 3 \).

Answer:

\( 2 \)

Problem 2