Solve For \( x \) \( \begin{array}{l}\text { 1. } 2^{x}=64 \quad \text { 10. } x^{\frac{1}{4}}=1 \\ \text { 2. } 3^{x}=27 \\ \text { 3. } 7 \cdot 3^{x}=217 \\ \text { 4. } 3 \cdot 9^{x}=27 \\ \text { 5. } 2^{-3 x}=2^{2 x-2} \\ \text { 6. } 8^{x+2}=16^{2 x-3}\end{array} \)

We have several equations to solve for x. Let’s solve each one step by step. 1. Equation: 2^x = 64 Rewrite 64 as a power of 2. Since 64 = 2^6, we have   2^x = 2^6 Therefore, by the one‐to‐one property, x = 6. 2. Equation: 3^x = 27 Notice that 27 = 3^3, so   3^x = 3^3 Thus, x = 3. 3. Equation: 7·3^x = 217 First, isolate 3^x by dividing both sides by 7:   3^x = 217 ÷ 7 = 31 Since 31 is not a power of 3, express x in logarithmic form:   x = log₃(31). 4. Equation: 3·9^x = 27 Express 9^x as (3^2)^x = 3^(2x). Then the equation becomes:   3 · 3^(2x) = 27 Combine the factors:   3^(2x + 1) = 27 Recognize that 27 = 3^3, so   3^(2x + 1) = 3^3 Therefore, 2x + 1 = 3, and solving gives:   2x = 2 ⟹ x = 1. 5. Equation: 2^(–3x) = 2^(2x – 2) Since the bases are the same, set the exponents equal:   –3x = 2x – 2 Solve for x:   –3x – 2x = –2 ⟹ –5x = –2 ⟹ x = 2/5. 6. Equation: 8^(x + 2) = 16^(2x – 3) Express both sides with base 2:   8 = 2³ and 16 = 2⁴, so   (2³)^(x + 2) = (2⁴)^(2x – 3) This becomes:   2^(3x + 6) = 2^(8x – 12) Equate the exponents:   3x + 6 = 8x – 12 Solve for x:   6 + 12 = 8x – 3x ⟹ 18 = 5x ⟹ x = 18/5. 10. Equation: x^(1/4) = 1 To eliminate the fourth root, raise both sides to the 4th power:   (x^(1/4))⁴ = 1⁴   x = 1. Summary of Answers:  1. x = 6  2. x = 3  3. x = log₃(31)  4. x = 1  5. x = 2/5  6. x = 18/5  10. x = 1