Question

b^x b^y = b^4; \frac{b^{2(x)} b^4}{b^{2(y)}} = b^{20}; x=\boxed{}; y=\boxed{}

Explanation:

Step1: Use exponent rule \(a^m \cdot a^n = a^{m + n}\) on \(b^x b^y\)

We know that when multiplying exponents with the same base, we add the exponents. So \(b^x b^y = b^{x + y}\), and we are given that \(b^x b^y = b^4\), so we get the equation:
\(x + y = 4\) --- (1)

Step2: Simplify the fraction \(\frac{b^{2x}b^4}{b^{2y}}\) using exponent rules

First, use the multiplication rule \(a^m \cdot a^n = a^{m + n}\) on the numerator \(b^{2x}b^4\), we get \(b^{2x + 4}\). Then, use the division rule \(a^m\div a^n=a^{m - n}\) for \(\frac{b^{2x + 4}}{b^{2y}}\), we get \(b^{(2x + 4)-2y}=b^{2x-2y + 4}\). We are given that this is equal to \(b^{20}\), so we can set the exponents equal:
\(2x-2y + 4=20\)
Simplify this equation by subtracting 4 from both sides:
\(2x-2y=16\)
Divide both sides by 2:
\(x - y = 8\) --- (2)

Step3: Solve the system of equations

We have the system of equations:
\(

$$\begin{cases}x + y = 4\\x - y = 8\end{cases}$$

\)
Add the two equations together:
\((x + y)+(x - y)=4 + 8\)
Simplify the left side: \(2x=12\), so \(x = 6\)
Substitute \(x = 6\) into equation (1): \(6 + y = 4\), then \(y=4 - 6=-2\)

Answer:

\(x = 6\), \(y=-2\)