Q. If \(3x – y = 12\), what is the value of \(\frac{8^x}{2^y}\)?

Q: What is the value of \frac{8^x}{2^y} given 3x - y = 12?

Rewrite 8^x=(2^3)^x=2^{3x}. Then \frac{8^x}{2^y}=2^{3x-y}=2^{12}=4096.

Q: Why rewrite 8 as a power of 2?

Rewriting as 8=2^3 makes bases equal so you can combine exponents: 8^x=(2^3)^x=2^{3x}, allowing 2^{3x}/2^y=2^{3x-y}.

Q: Can x and y be specific numbers that satisfy 3x-y=12?

Yes infinitely many. Solve y=3x-12. Example: x=4 gives y=0, and 8^4/2^0=4096.

Q: Does the result depend on the individual values of x and y or just on 3x-y?

Only on the combination 3x-y. Any pair with 3x-y=12 yields 2^{12}=4096, regardless of individual x,y.

Q: How would the answer change if the equation were 3x-y=k?

How would the answer change if the equation were 3x-y=k?

Q: If x and y are not integers, is 4096 still valid?

Yes. The identity 8^x/2^y=2^{3x-y} holds for real x,y; if 3x-y=12, the value is 2^{12}=4096 for all real solutions.

Answer

\( \frac{8^{x}}{2^{y}} = \frac{(2^{3})^{x}}{2^{y}} = 2^{3x-y} = 2^{12} = 4096. \)

Detailed Explanation

Identify a way to rewrite 8 as a power of 2. Since 8 = 2^3, apply the exponent rule \((a^m)^n = a^{mn}\) to rewrite \(8^x\).

\(8^x = (2^3)^x = 2^{3x}\)
Rewrite the given expression using the result from step 1:

\(\dfrac{8^x}{2^y} = \dfrac{2^{3x}}{2^y}\)
Use the exponent rule for division with the same base: \(\dfrac{a^m}{a^n} = a^{m-n}\). Apply it to combine the powers of 2.

\(\dfrac{2^{3x}}{2^y} = 2^{3x – y}\)
Use the given linear relation to substitute for the exponent. We are given \(3x – y = 12\). Substitute this into the exponent:

\(2^{3x – y} = 2^{12}\)
Evaluate \(2^{12}\). One convenient way: \(2^{10} = 1024\), therefore \(2^{12} = 2^{10} \times 2^2 = 1024 \times 4 = 4096\).
Final answer: 4096

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FAQs

What is the value of \( \frac{8^x}{2^y} \) given \(3x - y = 12\)?

Rewrite \(8^x=(2^3)^x=2^{3x}\). Then \( \frac{8^x}{2^y}=2^{3x-y}=2^{12}=4096\).

Why rewrite \(8\) as a power of \(2\)?

Rewriting as \(8=2^3\) makes bases equal so you can combine exponents: \(8^x=(2^3)^x=2^{3x}\), allowing \(2^{3x}/2^y=2^{3x-y}\).

Can \(x\) and \(y\) be specific numbers that satisfy \(3x-y=12\)?

Yes infinitely many. Solve \(y=3x-12\). Example: \(x=4\) gives \(y=0\), and \(8^4/2^0=4096\).

Does the result depend on the individual values of \(x\) and \(y\) or just on \(3x-y\)?

Only on the combination \(3x-y\). Any pair with \(3x-y=12\) yields \(2^{12}=4096\), regardless of individual \(x,y\).

What is the prime factorization of the result?

\(4096=2^{12}\).

How would the answer change if the equation were \(3x-y=k\)?

What common mistakes should I avoid?

Don’t divide exponents incorrectly; convert to same base first. Avoid doing \(8^x/2^y=4^{x-y}\) or subtracting bases; work with exponents: \(2^{3x-y}\).

If \(x\) and \(y\) are not integers, is \(4096\) still valid?

Yes. The identity \(8^x/2^y=2^{3x-y}\) holds for real \(x,y\); if \(3x-y=12\), the value is \(2^{12}=4096\) for all real solutions.

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