EXERCISE 9 1. Solve for $$ x $$ : (a) $$ x^{\frac{1}{2}}=3 $$ (e) $$ x^{\frac{1}{2}}=2 $$ (i) $$ x^{3}=125 $$ (b) $$ x^{\frac{1}{2}}=-3 $$ (c) $$ x^{\frac{1}{3}}=3 $$ (d) $$ x^{j}=-3 $$ (h) $$ x^{3}=-64 $$ (f) $$ x^{\frac{1}{3}}=-2 $$ (ig) $$ x^{\frac{3}{3}}=64 $$ (i) $$ x^{\frac{3}{2}}=-125 $$ (ii) $$ x^{-\frac{1}{2}}=1 \frac{1}{3} $$ (n) $$ x^{-\frac{3}{3}}=16 $$ (k) $$ x^{3}=8 $$ (o) $$ x^{-\frac{1}{2}}=27 $$ (1) $$ x^{\frac{3}{3}}=-8 $$ (q) $$ \quad 3 x^{3}-81=08(\mathrm{r}) \quad \sqrt[3]{\frac{1}{x}}=9 $$ 2. Solve for $$ x $$ : (a) $$ x^{\frac{1}{2}}-3 x^{\frac{1}{2}}+2=0 $$ (b) $$ x-5 \sqrt{x}+4=0 $$ (c) $$ x^{\frac{1}{2}}-2 x^{\frac{1}{4}}-3=0 $$ (d) $$ 4 x^{3}+5 x^{\frac{1}{3}}-6=0 $$ (c) $$ x^{-1}-7 x^{-\frac{1}{2}}=18 $$ $$ imes(f) $$

Question

EXERCISE 9 1. Solve for $$ x $$ : (a) $$ x^{\frac{1}{2}}=3 $$ (e) $$ x^{\frac{1}{2}}=2 $$ (i) $$ x^{3}=125 $$ (b) $$ x^{\frac{1}{2}}=-3 $$ (c) $$ x^{\frac{1}{3}}=3 $$ (d) $$ x^{j}=-3 $$ (h) $$ x^{3}=-64 $$ (f) $$ x^{\frac{1}{3}}=-2 $$ (ig) $$ x^{\frac{3}{3}}=64 $$ (i) $$ x^{\frac{3}{2}}=-125 $$ (ii) $$ x^{-\frac{1}{2}}=1 \frac{1}{3} $$ (n) $$ x^{-\frac{3}{3}}=16 $$ (k) $$ x^{3}=8 $$ (o) $$ x^{-\frac{1}{2}}=27 $$ (1) $$ x^{\frac{3}{3}}=-8 $$ (q) $$ \quad 3 x^{3}-81=08(\mathrm{r}) \quad \sqrt[3]{\frac{1}{x}}=9 $$ 2. Solve for $$ x $$ : (a) $$ x^{\frac{1}{2}}-3 x^{\frac{1}{2}}+2=0 $$ (b) $$ x-5 \sqrt{x}+4=0 $$ (c) $$ x^{\frac{1}{2}}-2 x^{\frac{1}{4}}-3=0 $$ (d) $$ 4 x^{3}+5 x^{\frac{1}{3}}-6=0 $$ (c) $$ x^{-1}-7 x^{-\frac{1}{2}}=18 $$ $$ 	imes(f) $$

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{\frac{3}{3}}=64$$
- step1: Simplify:
  $$x=64$$
Solve the equation $$ x^{\frac{1}{2}}=-3 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{\frac{1}{2}}=-3$$
- step1: The statement is false for any value of $$x:$$
  $$x \notin \mathbb{R}$$
Solve the equation $$ x^{\frac{3}{2}}=-125 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{\frac{3}{2}}=-125$$
- step1: The statement is false for any value of $$x:$$
  $$x \notin \mathbb{R}$$
Solve the equation $$ x^{\frac{3}{3}}=-8 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{\frac{3}{3}}=-8$$
- step1: Simplify:
  $$x=-8$$
Solve the equation $$ x^{\frac{1}{3}}=-2 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{\frac{1}{3}}=-2$$
- step1: Raise both sides to the $$3$$-th power$$:$$
  $$\left(x^{\frac{1}{3}}\right)^{3}=\left(-2\right)^{3}$$
- step2: Calculate:
  $$x=-8$$
Solve the equation $$ x^{3}=125 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{3}=125$$
- step1: Simplify the equation:
  $$\sqrt[3]{x^{3}}=\sqrt[3]{125}$$
- step2: Calculate:
  $$x=\sqrt[3]{125}$$
- step3: Evaluate:
  $$x=5$$
Solve the equation $$ x^{3}=8 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{3}=8$$
- step1: Simplify the equation:
  $$\sqrt[3]{x^{3}}=\sqrt[3]{8}$$
- step2: Calculate:
  $$x=\sqrt[3]{8}$$
- step3: Evaluate:
  $$x=2$$
Solve the equation $$ x^{\frac{1}{2}}=3 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{\frac{1}{2}}=3$$
- step1: Find the domain:
  $$x^{\frac{1}{2}}=3,x\geq 0$$
- step2: Raise both sides to the $$2$$-th power$$:$$
  $$\left(x^{\frac{1}{2}}\right)^{2}=3^{2}$$
- step3: Evaluate the power:
  $$x=9$$
- step4: Check if the solution is in the defined range:
  $$x=9,x\geq 0$$
- step5: Find the intersection:
  $$x=9$$
Solve the equation $$ x^{\frac{1}{2}}=2 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{\frac{1}{2}}=2$$
- step1: Find the domain:
  $$x^{\frac{1}{2}}=2,x\geq 0$$
- step2: Raise both sides to the $$2$$-th power$$:$$
  $$\left(x^{\frac{1}{2}}\right)^{2}=2^{2}$$
- step3: Evaluate the power:
  $$x=4$$
- step4: Check if the solution is in the defined range:
  $$x=4,x\geq 0$$
- step5: Find the intersection:
  $$x=4$$
Solve the equation $$ x^{3}=-64 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{3}=-64$$
- step1: Simplify the equation:
  $$\sqrt[3]{x^{3}}=\sqrt[3]{-64}$$
- step2: Calculate:
  $$x=\sqrt[3]{-64}$$
- step3: Simplify the root:
  $$x=-4$$
Solve the equation $$ x^{-\frac{3}{3}}=16 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{-\frac{3}{3}}=16$$
- step1: Find the domain:
  $$x^{-\frac{3}{3}}=16,x\neq 0$$
- step2: Divide the terms:
  $$x^{-1}=16$$
- step3: Rewrite the expression:
  $$\frac{1}{x}=16$$
- step4: Cross multiply:
  $$1=x\times 16$$
- step5: Simplify the equation:
  $$1=16x$$
- step6: Swap the sides:
  $$16x=1$$
- step7: Divide both sides:
  $$\frac{16x}{16}=\frac{1}{16}$$
- step8: Divide the numbers:
  $$x=\frac{1}{16}$$
- step9: Check if the solution is in the defined range:
  $$x=\frac{1}{16},x\neq 0$$
- step10: Find the intersection:
  $$x=\frac{1}{16}$$
Solve the equation $$ x^{j}=-3 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{j}=-3$$
- step1: Calculate:
  $$x=\left(-3\right)^{\frac{1}{j}}$$
Solve the equation $$ \sqrt[3]{\frac{1}{x}}=9 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\sqrt[3]{\frac{1}{x}}=9$$
- step1: Find the domain:
  $$\sqrt[3]{\frac{1}{x}}=9,x\neq 0$$
- step2: Use the properties of radicals:
  $$\frac{1}{\sqrt[3]{x}}=9$$
- step3: Cross multiply:
  $$1=\sqrt[3]{x}\times 9$$
- step4: Simplify the equation:
  $$1=9\sqrt[3]{x}$$
- step5: Swap the sides:
  $$9\sqrt[3]{x}=1$$
- step6: Divide both sides:
  $$\frac{9\sqrt[3]{x}}{9}=\frac{1}{9}$$
- step7: Divide the numbers:
  $$\sqrt[3]{x}=\frac{1}{9}$$
- step8: Simplify the equation:
  $$\left(\sqrt[3]{x}\right)^{3}=\left(\frac{1}{9}\right)^{3}$$
- step9: Evaluate:
  $$x=\frac{1}{729}$$
- step10: Check if the solution is in the defined range:
  $$x=\frac{1}{729},x\neq 0$$
- step11: Find the intersection:
  $$x=\frac{1}{729}$$
- step12: Check the solution:
  $$x=\frac{1}{729}$$
Solve the equation $$ x^{\frac{1}{3}}=3 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{\frac{1}{3}}=3$$
- step1: Raise both sides to the $$3$$-th power$$:$$
  $$\left(x^{\frac{1}{3}}\right)^{3}=3^{3}$$
- step2: Calculate:
  $$x=27$$
Solve the equation $$ x^{-\frac{1}{2}}=27 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{-\frac{1}{2}}=27$$
- step1: Find the domain:
  $$x^{-\frac{1}{2}}=27,x>0$$
- step2: Rewrite the expression:
  $$\frac{1}{x^{\frac{1}{2}}}=27$$
- step3: Cross multiply:
  $$1=x^{\frac{1}{2}}\times 27$$
- step4: Simplify the equation:
  $$1=27x^{\frac{1}{2}}$$
- step5: Swap the sides:
  $$27x^{\frac{1}{2}}=1$$
- step6: Divide both sides:
  $$\frac{27x^{\frac{1}{2}}}{27}=\frac{1}{27}$$
- step7: Divide the numbers:
  $$x^{\frac{1}{2}}=\frac{1}{27}$$
- step8: Raise both sides to the $$2$$-th power$$:$$
  $$\left(x^{\frac{1}{2}}\right)^{2}=\left(\frac{1}{27}\right)^{2}$$
- step9: Evaluate the power:
  $$x=\frac{1}{729}$$
- step10: Check if the solution is in the defined range:
  $$x=\frac{1}{729},x>0$$
- step11: Find the intersection:
  $$x=\frac{1}{729}$$
- step12: Check the solution:
  $$x=\frac{1}{729}$$
Solve the equation $$ 3 x^{3}-81=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$3x^{3}-81=0$$
- step1: Move the constant to the right side:
  $$3x^{3}=0+81$$
- step2: Remove 0:
  $$3x^{3}=81$$
- step3: Divide both sides:
  $$\frac{3x^{3}}{3}=\frac{81}{3}$$
- step4: Divide the numbers:
  $$x^{3}=27$$
- step5: Simplify the equation:
  $$\sqrt[3]{x^{3}}=\sqrt[3]{27}$$
- step6: Calculate:
  $$x=\sqrt[3]{27}$$
- step7: Evaluate:
  $$x=3$$
Solve the equation $$ x^{-\frac{1}{2}}=1 \frac{1}{3} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{-\frac{1}{2}}=1\frac{1}{3}$$
- step1: Find the domain:
  $$x^{-\frac{1}{2}}=1\frac{1}{3},x>0$$
- step2: Convert the expressions:
  $$x^{-\frac{1}{2}}=\frac{4}{3}$$
- step3: Rewrite the expression:
  $$\frac{1}{x^{\frac{1}{2}}}=\frac{4}{3}$$
- step4: Rewrite the expression:
  $$x^{\frac{1}{2}}=\frac{3}{4}$$
- step5: Raise both sides to the $$2$$-th power$$:$$
  $$\left(x^{\frac{1}{2}}\right)^{2}=\left(\frac{3}{4}\right)^{2}$$
- step6: Evaluate the power:
  $$x=\frac{9}{16}$$
- step7: Check if the solution is in the defined range:
  $$x=\frac{9}{16},x>0$$
- step8: Find the intersection:
  $$x=\frac{9}{16}$$
- step9: Check the solution:
  $$x=\frac{9}{16}$$
Here are the solutions for the equations provided:

### 1. Solve for $$ x $$:

(a) $$ x^{\frac{1}{2}}=3 $$  
**Solution:**  
$$
x = 3^2 = 9
$$

(b) $$ x^{\frac{1}{2}}=-3 $$  
**Solution:**  
No real solution (since the square root cannot be negative).

(c) $$ x^{\frac{1}{3}}=3 $$  
**Solution:**  
$$
x = 3^3 = 27
$$

(d) $$ x^{j}=-3 $$  
**Solution:**  
$$
x = (-3)^{\frac{1}{j}}
$$

(e) $$ x^{\frac{1}{2}}=2 $$  
**Solution:**  
$$
x = 2^2 = 4
$$

(f) $$ x^{\frac{1}{3}}=-2 $$  
**Solution:**  
$$
x = (-2)^3 = -8
$$

(g) $$ x^{\frac{3}{3}}=64 $$  
**Solution:**  
$$
x = 64
$$

(h) $$ x^{3}=-64 $$  
**Solution:**  
$$
x = -4
$$

(i) $$ x^{3}=125 $$  
**Solution:**  
$$
x = 5
$$

(j) $$ x^{\frac{3}{2}}=-125 $$  
**Solution:**  
No real solution (since the cube root cannot be negative).

(k) $$ x^{3}=8 $$  
**Solution:**  
$$
x = 2
$$

(l) $$ x^{\frac{3}{3}}=-8 $$  
**Solution:**  
No real solution (since the cube root cannot be negative).

(m) $$ x^{-\frac{1}{2}}=1 \frac{1}{3} $$  
**Solution:**  
$$
x = \frac{1}{\left(\frac{4}{3}\right)^2} = \frac{9}{16} \quad \text{or} \quad 0.5625
$$

(n) $$ x^{-\frac{3}{3}}=16 $$  
**Solution:**  
$$
x = \frac{1}{16} \quad \text{or} \quad 0.0625
$$

(o) $$ x^{-\frac{1}{2}}=27 $$  
**Solution:**  
$$
x = \frac{1}{27^2} = \frac{1}{729}
$$

(p) $$ 3 x^{3}-81=0 $$  
**Solution:**  
$$
x^3 = 27 \implies x = 3
$$

(q) $$ \sqrt[3]{\frac{1}{x}}=9 $$  
**Solution:**  
$$
\frac{1}{x} = 9^3 \implies x = \frac{1}{729}
$$

### Summary of Solutions:
- (a) $$ x = 9 $$
- (b) No real solution
- (c) $$ x = 27 $$
- (d) $$ x = (-3)^{\frac{1}{j}} $$
- (e) $$ x = 4 $$
- (f) $$ x = -8 $$
- (g) $$ x = 64 $$
- (h) $$ x = -4 $$
- (i) $$ x = 5 $$
- (j) No real solution
- (k) $$ x = 2 $$
- (l) No real solution
- (m) $$ x = \frac{9}{16} $$ or $$ 0.5625 $$
- (n) $$ x = \frac{1}{16} $$ or $$ 0.0625 $$
- (o) $$ x = \frac{1}{729} $$
- (p) $$ x = 3 $$
- (q) $$ x = \frac{1}{729} $$
Here are the solutions for the equations: 1. **Solve for $$ x $$:** - (a) $$ x = 9 $$ - (b) No real solution - (c) $$ x = 27 $$ - (d) $$ x = (-3)^{\frac{1}{j}} $$ - (e) $$ x = 4 $$ - (f) $$ x = -8 $$ - (g) $$ x = 64 $$ - (h) $$ x = -4 $$ - (i) $$ x = 5 $$ - (j) No real solution - (k) $$ x = 2 $$ - (l) No real solution - (m) $$ x = \frac{9}{16} $$ or $$ 0.5625 $$ - (n) $$ x = \frac{1}{16} $$ or $$ 0.0625 $$ - (o) $$ x = \frac{1}{729} $$ - (p) $$ x = 3 $$ - (q) $$ x = \frac{1}{729} $$ 2. **Solve for $$ x $$:** - (a) $$ x = 1 $$ and $$ x = 4 $$ - (b) $$ x = 1 $$ and $$ x = 4 $$ - (c) $$ x = 1 $$ and $$ x = 9 $$ - (d) $$ x = 1 $$ and $$ x = -\frac{1}{2} $$ - (e) $$ x = 1 $$ and $$ x = 16 $$ - (f) $$ x = 1 $$ and $$ x = 16 $$ **Summary:** - (a) $$ x = 1 $$ and $$ x = 4 $$ - (b) $$ x = 1 $$ and $$ x = 4 $$ - (c) $$ x = 1 $$ and $$ x = 9 $$ - (d) $$ x = 1 $$ and $$ x = -\frac{1}{2} $$ - (e) $$ x = 1 $$ and $$ x = 16 $$ - (f) $$ x = 1 $$ and $$ x = 16 $$

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