Question

name nobel
quiz 6
solve each exponent.

1. $18^x$
2. $t^6$
3. $\left(\frac{9}{1}\

ight)^4$

1. $5^{n - 3}$
2. $p^{-4}$
3. $10^7$

write each radical using rational exponents.

1. $\sqrt{14}$
2. $\sqrtk{12^3}$
3. $\sqrt8{11^8}$
4. $\sqrt3{\sqrt{v^8}}$
5. $\sqrt{u^9}$
6. $\sqrtk{j^d}$

write each rational exponent as a radical.

1. $10^{\frac{1}{5}}$
2. $10^{\frac{y}{5}}$
3. $c^{\frac{h}{t}}$

solve each equation:

1. $(2^{3x + 2})(2^{5x - 1}) = 2^{25}$
2. $6^{x - 8} = 36^{(2x - 7)}$
3. $\frac{27^{\frac{2x}{3}}}{27^{\frac{x}{3}}} = 3^3$

Explanation:

Step1: Solve \(18^x\) (assuming "solve" here means expand or analyze, but since it's an exponential, maybe rewrite? Wait, the first part says "Solve each exponent" – maybe evaluate or rewrite. Wait, maybe the first problem is to write in expanded form? Wait, \(18^x\) is \(18 \times 18 \times \dots \times 18\) (x times). But maybe the task is to simplify or rewrite. Wait, let's take problem 16 first as an example.

Step1 (Problem 16): Use exponent rule \(a^m \times a^n = a^{m+n}\)

Given \((2^{3x + 2})(2^{5x - 1}) = 2^{25}\), apply the rule: \(2^{(3x + 2) + (5x - 1)} = 2^{25}\)
Simplify exponent: \(3x + 2 + 5x - 1 = 8x + 1\), so \(2^{8x + 1} = 2^{25}\)

Step2: Set exponents equal (since bases are equal)

\(8x + 1 = 25\)
Subtract 1: \(8x = 24\)
Divide by 8: \(x = 3\)

Answer:

(for Problem 16):
\(x = 3\)