63. write the exponential expression ( 3x^{\frac{3}{8}} ) in radical form.
a. ( 3sqrt8{x^3} ) b. ( sqrt8{3x^3} ) c. ( 3sqrt3{x^8} ) d. ( 3^{\frac{3}{8}}sqrt8{x^3} )
64. write the radical expression ( \frac{8}{sqrt7{x^{13}}} ) in exponential form.
a. ( 8x^{\frac{7}{13}} ) b. ( 8x^{\frac{13}{7}} ) c. ( 8x^{-\frac{13}{7}} ) d. ( 8x^{\frac{7}{13}} )
simplify.
65. ( 16^{\frac{1}{2}} )
a. ( 16^2 ) b. ( 4 ) c. ( sqrt{16^2} ) d. ( 16 )
66. what is ( \frac{sqrt3{x^3}}{sqrt5{x^2}} ) in simplest form?
a. ( x^{\frac{3}{5}} ) b. ( x^{\frac{5}{3}} ) c. ( x^{\frac{9}{15}} ) d. ( x^{\frac{15}{9}} )
67. ( \frac{sqrt{90x^{18}}}{sqrt{2x}} )
a. ( 3x^8sqrt{5x} ) b. ( sqrt{18x^{17}} ) c. ( 5xsqrt{3x^8} ) d. none of these
what is the solution of the equation? eliminate any extraneous solutions.
68. ( 5x = sqrt{10 + 15x} )
a. ( -1 ) b. ( 1 ) and ( -\frac{2}{5} ) c. ( 1 ) d. ( -\frac{2}{5} )
what is the solution of the equation?
69. ( sqrt{x + 10} - 7 = -5 )
a. ( 14 ) b. ( -8 ) c. ( 4 ) d. ( -6 )
70. ( -10 + sqrt{x + 8} = -4 )
a. ( 36 ) b. ( 28 ) c. ( -2 ) d. ( 44 )

Question

63. write the exponential expression ( 3x^{\frac{3}{8}} ) in radical form.
 a. ( 3sqrt8{x^3} )  b. ( sqrt8{3x^3} )  c. ( 3sqrt3{x^8} )  d. ( 3^{\frac{3}{8}}sqrt8{x^3} )
64. write the radical expression ( \frac{8}{sqrt7{x^{13}}} ) in exponential form.
 a. ( 8x^{\frac{7}{13}} )  b. ( 8x^{\frac{13}{7}} )  c. ( 8x^{-\frac{13}{7}} )  d. ( 8x^{\frac{7}{13}} )
simplify.
65. ( 16^{\frac{1}{2}} )
 a. ( 16^2 )  b. ( 4 )  c. ( sqrt{16^2} )  d. ( 16 )
66. what is ( \frac{sqrt3{x^3}}{sqrt5{x^2}} ) in simplest form?
 a. ( x^{\frac{3}{5}} )  b. ( x^{\frac{5}{3}} )  c. ( x^{\frac{9}{15}} )  d. ( x^{\frac{15}{9}} )
67. ( \frac{sqrt{90x^{18}}}{sqrt{2x}} )
 a. ( 3x^8sqrt{5x} )  b. ( sqrt{18x^{17}} )  c. ( 5xsqrt{3x^8} )  d. none of these
what is the solution of the equation? eliminate any extraneous solutions.
68. ( 5x = sqrt{10 + 15x} )
 a. ( -1 )  b. ( 1 ) and ( -\frac{2}{5} )  c. ( 1 )  d. ( -\frac{2}{5} )
what is the solution of the equation?
69. ( sqrt{x + 10} - 7 = -5 )
 a. ( 14 )  b. ( -8 )  c. ( 4 )  d. ( -6 )
70. ( -10 + sqrt{x + 8} = -4 )
 a. ( 36 )  b. ( 28 )  c. ( -2 )  d. ( 44 )

Accepted Answer

### Question 63
# Explanation:
## Step1: Recall the rule $ a x^{\frac{m}{n}} = a \sqrt[n]{x^m} $
Given $ 3x^{\frac{3}{8}} $, here $ a = 3 $, $ m = 3 $, $ n = 8 $
## Step2: Apply the rule
So $ 3x^{\frac{3}{8}} = 3\sqrt[8]{x^3} $ (Wait, no, wait the options: option a is $ 3\sqrt[8]{x^3} $? Wait the original problem: the exponential expression is $ 3x^{\frac{3}{8}} $? Wait the user's image: "Write the exponential expression $ 3x^{\frac{3}{8}} $ in radical form." Wait the options: a. $ 3\sqrt[8]{x^3} $, b. $ \sqrt[8]{3x^3} $, c. $ 3\sqrt[8]{x^8} $, d. $ 3^{\frac{3}{8}}\sqrt[8]{x^3} $. Wait, the correct rule is $ x^{\frac{m}{n}}=\sqrt[n]{x^m} $, so $ 3x^{\frac{3}{8}} = 3\times\sqrt[8]{x^3} $, which is option a? Wait no, maybe I misread. Wait the exponent is $ \frac{3}{8} $, so the radical is 8th root, and the power of x is 3. So $ 3x^{\frac{3}{8}} = 3\sqrt[8]{x^3} $, which is option a.
# Answer: a. $ 3\sqrt[8]{x^3} $

### Question 64
# Explanation:
## Step1: Recall the rule for radicals in denominator: $ \frac{a}{\sqrt[n]{x^m}} = a x^{-\frac{m}{n}} $
Given $ \frac{8}{\sqrt[7]{x^{13}}} $, here $ a = 8 $, $ n = 7 $, $ m = 13 $
## Step2: Apply the rule
So $ \frac{8}{\sqrt[7]{x^{13}}} = 8x^{-\frac{13}{7}} $? Wait no, wait the options: a. $ 8x^{\frac{7}{13}} $, b. $ 8x^{\frac{13}{7}} $, c. $ 8x^{-\frac{13}{7}} $, d. $ 8x^{\frac{7}{13}} $. Wait, the rule is $ \sqrt[n]{x^m}=x^{\frac{m}{n}} $, so $ \frac{1}{\sqrt[7]{x^{13}}}=\frac{1}{x^{\frac{13}{7}}}=x^{-\frac{13}{7}} $, so $ \frac{8}{\sqrt[7]{x^{13}}}=8x^{-\frac{13}{7}} $, which is option c.
# Answer: c. $ 8x^{-\frac{13}{7}} $

### Question 65
# Explanation:
## Step1: Recall $ a^{\frac{1}{2}}=\sqrt{a} $
So $ 16^{\frac{1}{2}}=\sqrt{16} $
## Step2: Calculate $ \sqrt{16} $
$ \sqrt{16}=4 $
# Answer: b. 4

### Question 66
# Explanation:
## Step1: Convert radicals to exponents: $ \sqrt[3]{x^3}=x^{\frac{3}{3}}=x^1 $? Wait no, $ \sqrt[3]{x^3}=x^{\frac{3}{3}}=x^1 $, and $ \sqrt[5]{x^2}=x^{\frac{2}{5}} $. Wait the expression is $ \frac{\sqrt[3]{x^3}}{\sqrt[5]{x^2}} $
## Step2: Apply exponent rules: $ \frac{x^a}{x^b}=x^{a - b} $
First, $ \sqrt[3]{x^3}=x^{\frac{3}{3}} = x^1 $, $ \sqrt[5]{x^2}=x^{\frac{2}{5}} $
So $ \frac{x^1}{x^{\frac{2}{5}}}=x^{1-\frac{2}{5}}=x^{\frac{3}{5}} $? Wait no, wait the options: a. $ x^{\frac{3}{5}} $, b. $ x^{\frac{5}{3}} $, c. $ x^{\frac{9}{15}} $ (which is $ x^{\frac{3}{5}} $), d. $ x^{\frac{15}{9}} $. Wait, maybe I misread the radicals. Wait the numerator is $ \sqrt[3]{x^3} $? Wait no, maybe the numerator is $ \sqrt[3]{x^3} $? Wait no, the original problem: "What is $ \frac{\sqrt[3]{x^3}}{\sqrt[5]{x^2}} $ in simplest form?" Wait no, maybe the numerator is $ \sqrt[3]{x^3} $? Wait no, $ \sqrt[3]{x^3}=x $, $ \sqrt[5]{x^2}=x^{\frac{2}{5}} $, so $ \frac{x}{x^{\frac{2}{5}}}=x^{1 - \frac{2}{5}}=x^{\frac{3}{5}} $, which is option a. Wait but option c is $ x^{\frac{9}{15}} $, which is equal to $ x^{\frac{3}{5}} $ (since 9/15=3/5). Wait maybe the radicals are different. Wait maybe the numerator is $ \sqrt[3]{x^3} $? Wait no, perhaps the numerator is $ \sqrt[3]{x^3} $ and denominator $ \sqrt[5]{x^2} $. So the correct answer is a or c? Wait 3/5 is 9/15? No, 3/5=9/15? 3*3=9, 5*3=15, yes. So $ x^{\frac{3}{5}} = x^{\frac{9}{15}} $? Wait no, 3/5 is 0.6, 9/15 is 0.6. So both a and c? But that can't be. Wait maybe the original problem is $ \frac{\sqrt[3]{x^3}}{\sqrt[5]{x^2}} $? Wait no, maybe the numerator is $ \sqrt[3]{x^3} $ and denominator $ \sqrt[5]{x^2} $. Wait, let's re-express:

$ \sqrt[3]{x^3} = x^{\frac{3}{3}} = x^1 $

$ \sqrt[5]{x^2} = x^{\frac{2}{5}} $

So $ \frac{x^1}{x^{\frac{2}{5}}} = x^{1 - \frac{2}{5}} = x^{\frac{3}{5}} $, which is option a. Option c is $ x^{\frac{9}{15}} $, which is the same as $ x^{\frac{3}{5}} $ (since 9÷3=3, 15÷3=5). So maybe the problem has a typo, but the correct exponent is $ \frac{3}{5} $, which is option a.
# Answer: a. $ x^{\frac{3}{5}} $

### Question 67
# Explanation:
## Step1: Recall $ \frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}} $
So $ \frac{\sqrt{90x^{18}}}{\sqrt{2x}} = \sqrt{\frac{90x^{18}}{2x}} $
## Step2: Simplify the fraction inside the radical
$ \frac{90x^{18}}{2x} = 45x^{17} $ (Wait 90÷2=45, $ x^{18}÷x = x^{17} $)
Wait no, 90x^18 divided by 2x: 90/2=45, x^18/x = x^(18-1)=x^17. So $ \sqrt{45x^{17}} $
## Step3: Simplify $ \sqrt{45x^{17}} $
45=9*5, $ x^{17}=x^{16}*x $
So $ \sqrt{9*5*x^{16}*x} = \sqrt{9}*\sqrt{x^{16}}*\sqrt{5x} = 3x^8\sqrt{5x} $, which is option a.
# Answer: a. $ 3x^8\sqrt{5x} $

### Question 68
# Explanation:
## Step1: Square both sides to eliminate the square root
$ (5x)^2 = (\sqrt{10 + 15x})^2 $
## Step2: Simplify both sides
Left: $ 25x^2 $, Right: $ 10 + 15x $
So $ 25x^2 - 15x - 10 = 0 $
## Step3: Divide by 5
$ 5x^2 - 3x - 2 = 0 $
## Step4: Factor the quadratic
$ 5x^2 - 5x + 2x - 2 = 0 $ → $ 5x(x - 1) + 2(x - 1) = 0 $ → $ (5x + 2)(x - 1) = 0 $
## Step5: Solve for x
$ 5x + 2 = 0 $ → $ x = -\frac{2}{5} $; $ x - 1 = 0 $ → $ x = 1 $
## Step6: Check for extraneous solutions
Check $ x = 1 $: Left: 5(1)=5, Right: $ \sqrt{10 + 15(1)} = \sqrt{25}=5 $. Valid.

Check $ x = -\frac{2}{5} $: Left: 5*(-2/5)=-2, Right: $ \sqrt{10 + 15*(-2/5)} = \sqrt{10 - 6} = \sqrt{4}=2 $. Left is -2, Right is 2. Not equal. So extraneous.

So only x=1 is valid.
# Answer: c. 1

### Question 69
# Explanation:
## Step1: Isolate the square root
$ \sqrt{x + 10} - 7 = -5 $ → $ \sqrt{x + 10} = -5 + 7 = 2 $
## Step2: Square both sides
$ (\sqrt{x + 10})^2 = 2^2 $ → $ x + 10 = 4 $
## Step3: Solve for x
$ x = 4 - 10 = -6 $? Wait no, 4 - 10 = -6? Wait no, 2 squared is 4, so x + 10 = 4 → x = 4 - 10 = -6? But wait the options: a.14, b.-8, c.4, d.-6. Wait let's check:

If x=-6: $ \sqrt{-6 + 10} -7 = \sqrt{4} -7 = 2 -7 = -5 $, which matches the equation. Wait but earlier calculation: $ \sqrt{x + 10} = 2 $, so x + 10 = 4 → x = -6. So option d. Wait but I made a mistake earlier. Let's redo:

Step1: $ \sqrt{x + 10} - 7 = -5 $

Add 7 to both sides: $ \sqrt{x + 10} = 2 $

Square both sides: $ x + 10 = 4 $

Subtract 10: $ x = 4 - 10 = -6 $. So option d.
# Answer: d. $ -\frac{2}{5} $? No, question 69: answer is d. -6.

### Question 70
# Explanation:
## Step1: Isolate the square root
$ -10 + \sqrt{x + 8} = -4 $ → $ \sqrt{x + 8} = -4 + 10 = 6 $
## Step2: Square both sides
$ (\sqrt{x + 8})^2 = 6^2 $ → $ x + 8 = 36 $
## Step3: Solve for x
$ x = 36 - 8 = 28 $? Wait no, 36 - 8 = 28? Wait options: a.36, b.28, c.-2, d.44. Wait:

If x=28: $ -10 + \sqrt{28 + 8} = -10 + \sqrt{36} = -10 + 6 = -4 $, which matches. Wait, wait: $ \sqrt{28 + 8} = \sqrt{36} = 6 $, so -10 + 6 = -4. Correct. So x=28, which is option b.
# Answer: b. 28

(Note: Due to the complexity of the image and potential misreads, the above solutions are based on standard radical and exponential rules. Each question was solved by applying the appropriate algebraic rules for exponents and radicals, then verifying solutions for extraneous roots where necessary.)

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QUESTION IMAGE

Question

Explanation:

Question 63

Step1: Recall the rule \( a x^{\frac{m}{n}} = a \sqrt[n]{x^m} \)

Step2: Apply the rule

Step1: Recall the rule for radicals in denominator: \( \frac{a}{\sqrt[n]{x^m}} = a x^{-\frac{m}{n}} \)

Step2: Apply the rule

Step1: Recall \( a^{\frac{1}{2}}=\sqrt{a} \)

Step2: Calculate \( \sqrt{16} \)

Answer:

Question 64