d) $$ \frac{2^{a+1} \cdot 3^{a-1}}{6^{a}} $$ g) $$ \frac{2^{1+2 n} \cdot 8^{1-3 n}}{4^{n+1}} $$ j) $$ \frac{(x y z)^{o}}{2^{-3}} $$

Question

UpStudy AI · Accepted Answer

We are given three expressions to simplify. Let’s work through each one step‐by‐step.

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(d) Simplify  (2^(a+1) · 3^(a–1)) / (6^a)

Step 1. Write 6^a as (2·3)^a = 2^a · 3^a.

Step 2. Substitute into the original expression:
  (2^(a+1) · 3^(a–1)) / (2^a · 3^a)

Step 3. Simplify the powers of 2:
  2^(a+1) / 2^a = 2^( (a+1) – a) = 2^1 = 2

Step 4. Simplify the powers of 3:
  3^(a–1) / 3^a = 3^( (a–1) – a) = 3^(–1) = 1/3

Step 5. Multiply the simplified parts:
  2 · (1/3) = 2/3

So, answer (d): 2/3

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(g) Simplify  (2^(1+2n) · 8^(1–3n)) / (4^(n+1))

Step 1. Express 8 and 4 as powers of 2:
  8 = 2^3  and  4 = 2^2

Step 2. Write 8^(1–3n) as (2^3)^(1–3n) = 2^(3(1–3n)) = 2^(3 – 9n).

Step 3. Write 4^(n+1) as (2^2)^(n+1) = 2^(2n+2).

Step 4. Now substitute these back:
  (2^(1+2n) · 2^(3–9n)) / 2^(2n+2)

Step 5. Combine the exponents in the numerator:
  2^( (1+2n)+(3–9n) ) = 2^(4 – 7n)

Step 6. Now the expression is:
  2^(4 – 7n) / 2^(2n+2) = 2^( (4 – 7n) – (2n + 2) )

Step 7. Simplify the exponent:
  (4 – 7n) – (2n + 2) = 4 – 7n – 2n – 2 = 2 – 9n

So, answer (g): 2^(2–9n)

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(j) Simplify  ((x y z)^o) / (2^(–3))

Note: Here the exponent “o” is treated as a variable (not the number zero).

Step 1. Recognize that dividing by 2^(–3) is the same as multiplying by 2^3:
  ((x y z)^o) / (2^(–3)) = (x y z)^o · 2^3

Step 2. Since 2^3 = 8, we have:
  8 · (x y z)^o

So, answer (j): 8·(xyz)^o

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Summary of Answers:

(d) 2/3  
(g) 2^(2–9n)  
(j) 8·(xyz)^o
(d) 2/3 (g) 2^(2–9n) (j) 8·(xyz)^o

d) \( \frac{2^{a+1} \cdot 3^{a-1}}{6^{a}} \) g) \( \frac{2^{1+2 n} \cdot 8^{1-3 n}}{4^{n+1}} \) j) \( \frac{(x y z)^{o}}{2^{-3}} \)