Question

1. simplify each to a power that has positive exponents only then evaluate.

(a) $16^{\frac{3}{4}}$
(b) $27^{-\frac{1}{3}}$
(4 mks.)

1. simplify fully: $\frac{(9a^{3}c)^{4}}{3c^{2}}$

(2 mks.)

1. solve the following equation. first get the bases the same on both sides.

$6^{2x - 4} = 216^{x - 3}$
(4 mks.)

end.
unit 6 quiz

Explanation:

Part 3(a): Simplify \( 16^{\frac{3}{4}} \)

Step 1: Rewrite 16 as a power of 2

We know that \( 16 = 2^4 \), so we can rewrite the expression as \( (2^4)^{\frac{3}{4}} \).

Step 2: Apply the exponent rule \( (a^m)^n = a^{m \times n} \)

Using the rule, we get \( 2^{4 \times \frac{3}{4}} \). The 4 in the numerator and denominator cancels out, leaving \( 2^3 \).

Step 3: Evaluate \( 2^3 \)

\( 2^3 = 8 \).

Step 1: Rewrite 27 as a power of 3

We know that \( 27 = 3^3 \), so we can rewrite the expression as \( (3^3)^{-\frac{1}{3}} \).

Step 2: Apply the exponent rule \( (a^m)^n = a^{m \times n} \)

Using the rule, we get \( 3^{3 \times (-\frac{1}{3})} \). The 3 in the numerator and denominator cancels out, leaving \( 3^{-1} \).

Step 3: Rewrite with positive exponent

Recall that \( a^{-n} = \frac{1}{a^n} \), so \( 3^{-1} = \frac{1}{3} \).

Step 1: Expand the numerator using \( (ab)^n = a^n b^n \)

Expanding \( (9a^3c)^4 \), we get \( 9^4 (a^3)^4 c^4 \). Calculating \( 9^4 = 6561 \), \( (a^3)^4 = a^{12} \), so the numerator becomes \( 6561a^{12}c^4 \).

Step 2: Divide by the denominator

We have \( \frac{6561a^{12}c^4}{3c^2} \). Divide the coefficients: \( \frac{6561}{3} = 2187 \). For the variables, use \( \frac{c^m}{c^n} = c^{m - n} \), so \( c^4 / c^2 = c^{4 - 2} = c^2 \).

Step 3: Combine the results

Putting it together, we get \( 2187a^{12}c^2 \).

Answer:

\( 8 \)

Part 3(b): Simplify \( 27^{-\frac{1}{3}} \)