Solve for $$ x $$ : $$ \begin{array}{l}2.2 .1 \quad 2-16 x^{-1}=0 \ 2.2 .2 \quad 4^{x}+8=9.2^{x} \ 2.2 .3 \quad \sqrt[x]{9}=243 \ ext { Simplify fully: } \ \frac{\sqrt{p^{2}-q^{2}} imes(p+q)^{\frac{5}{2}}}{(p-q)^{\frac{1}{2}}} \quad ext { if } p
eq q\end{array} $$

Question

Solve for $$ x $$ : $$ \begin{array}{l}2.2 .1 \quad 2-16 x^{-1}=0 \ 2.2 .2 \quad 4^{x}+8=9.2^{x} \ 2.2 .3 \quad \sqrt[x]{9}=243 \ 	ext { Simplify fully: } \ \frac{\sqrt{p^{2}-q^{2}} 	imes(p+q)^{\frac{5}{2}}}{(p-q)^{\frac{1}{2}}} \quad 	ext { if } p 
eq q\end{array} $$

UpStudy AI · Accepted Answer

1. Solve for $$ x $$ in $$ 2 - 16x^{-1} = 0 $$:
   
   - Rewrite the equation:
     $$
     2 - \frac{16}{x} = 0
     $$
   - Isolate the fraction:
     $$
     2 = \frac{16}{x}
     $$
   - Multiply both sides by $$ x $$:
     $$
     2x = 16
     $$
   - Divide both sides by 2:
     $$
     x = 8
     $$
   
2. Solve for $$ x $$ in $$ 4^{x}+8=9\cdot2^{x} $$:
   
   - Express $$ 4^{x} $$ as $$ (2^{2})^{x} = 2^{2x} $$. The equation becomes:
     $$
     2^{2x} + 8 = 9\cdot2^{x}
     $$
   - Let $$ y = 2^{x} $$. Then the equation becomes:
     $$
     y^{2} + 8 = 9y
     $$
   - Rearrange to form a quadratic equation:
     $$
     y^{2} - 9y + 8 = 0
     $$
   - Factor the quadratic:
     $$
     (y-1)(y-8) = 0
     $$
   - Solve for $$ y $$:
     $$
     y = 1 \quad \text{or} \quad y = 8
     $$
   - Convert back to $$ x $$:
     - For $$ y = 1 $$:
       $$
       2^{x} = 1 \quad \Rightarrow \quad x = 0
       $$
     - For $$ y = 8 $$:
       $$
       2^{x} = 8 \quad \Rightarrow \quad 2^{x} = 2^{3} \quad \Rightarrow \quad x = 3
       $$
   
3. Solve for $$ x $$ in $$ \sqrt[x]{9}=243 $$:
   
   - Express the radical as an exponent:
     $$
     9^{\frac{1}{x}} = 243
     $$
   - Write both sides as powers of 3:
     $$
     9 = 3^{2} \quad \text{and} \quad 243 = 3^{5}
     $$
   - Substitute to obtain:
     $$
     \left(3^{2}\right)^{\frac{1}{x}} = 3^{5} \quad \Rightarrow \quad 3^{\frac{2}{x}} = 3^{5}
     $$
   - Equate exponents:
     $$
     \frac{2}{x} = 5
     $$
   - Solve for $$ x $$:
     $$
     x = \frac{2}{5}
     $$
   
4. Simplify the expression
   $$
   \frac{\sqrt{p^{2}-q^{2}}\, (p+q)^{\frac{5}{2}}}{(p-q)^{\frac{1}{2}}}
   $$
   if $$ p \neq q $$:
   
   - Factor the expression under the square root:
     $$
     p^{2}-q^{2} = (p+q)(p-q)
     $$
   - Take the square root:
     $$
     \sqrt{p^{2}-q^{2}} = \sqrt{(p+q)(p-q)} = \sqrt{p+q}\sqrt{p-q}
     $$
   - Substitute back into the original expression:
     $$
     \frac{\sqrt{p+q}\sqrt{p-q}\, (p+q)^{\frac{5}{2}}}{(p-q)^{\frac{1}{2}}}
     $$
   - Combine like terms for $$(p+q)$$:
     $$
     \sqrt{p+q}\,(p+q)^{\frac{5}{2}} = (p+q)^{\frac{1}{2}+\frac{5}{2}} = (p+q)^{3}
     $$
   - Combine the radical terms for $$(p-q)$$:
     $$
     \frac{\sqrt{p-q}}{(p-q)^{\frac{1}{2}}} = 1
     $$
   - The simplified expression is:
     $$
     (p+q)^{3}
     $$

Final Answers:

1. $$ x = 8 $$

2. $$ x = 0 $$ or $$ x = 3 $$

3. $$ x = \frac{2}{5} $$

4. $$ (p+q)^{3} $$
$$ x = 8 $$, $$ x = 0 $$ or $$ x = 3 $$, $$ x = \frac{2}{5} $$, and the expression simplifies to $$ (p+q)^{3} $$.

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