Question

Which value is equivalent to 8 multiplied by 4 multiplied by 2 whole over 8 multiplied by 7, the whole raised to the power of 2 multiplied by 8 to the power of 0 over 7 to the power of negative 3, whole to the power of 3 multiplied by 7 to the power of negative 9?
64 over 49

8 over 49

16 over 7

512 over 7

Answer 1

The value is equivalent to 64 over 49

Further explanation

Let's recall following formula about Exponents and Surds:

$\boxed { \sqrt { x } = x ^ { \frac{1}{2} } }$

$\boxed { (a ^ b) ^ c = a ^ { b . c } }$

$\boxed {a ^ b \div a ^ c = a ^ { b - c } }$

$\boxed {\log a + \log b = \log (a \times b) }$

$\boxed {\log a - \log b = \log (a \div b) }$

Let us tackle the problem.

First, we will change this "word problem" into mathematical equation.

8 multiplied by 4 multiplied by 2 ↓

$8 \times 4 \times 2 = 64$

The whole over 8 multiplied by 7 ↓

$64 \div (8 \times 7) = 64 \div 56 = 8 \div 7$

The whole raised to the power of 2 ↓

$( 8 \div 7 )^2$

The whole multiplied by 8 to the power of 0 over 7 to the power of negative 3, whole to the power of 3 ↓

$(\frac{8}{7})^2 \times (\frac{8^0}{7^{-3}})^3$

$(\frac{8}{7})^2 \times (\frac{1}{7^{-9}})$

$(\frac{8^2}{7^2}) \times (\frac{1}{7^{-9}})$

$\frac{8^2}{7^{-7}}$

The whole multiplied by 7 to the power of negative 9 ↓

$\frac{8^2}{7^{-7}} \times 7^{-9}$

$\frac{8^2}{7^2}$

$\large {\boxed {\frac{64}{49} } }$

Learn more

• Coefficient of A Square Root : brainly.com/question/11337634
• The Order of Operations : brainly.com/question/10821615
• Write 100,000 Using Exponents : brainly.com/question/2032116

Answer details

Grade: High School

Subject: Mathematics

Chapter: Exponents and Surds

Keywords: Power , Multiplication , Division , Exponent , Surd , Negative , Postive , Value , Equivalent

Answer 2

Answer: First Option is correct.

Step-by-step explanation:

Since we have given that

8 multiplied by 4 multiplied by 2 whole over 8 multiplied by 7, the whole raised to the power of 2 multiplied by 8 to the power of 0 over 7 to the power of negative 3, whole to the power of 3 multiplied by 7 to the power of negative 9.

It means

$(\frac{8\times 4\times 2}{8\times 7})^{2} \times (\frac{8^0}{7^{-3}})^3 \times 7^{-9}$

So, when we cancel out 8 from numerator and denominator form the first expression, we get

$(\frac{ 4\times 2}{7})^{2} \times (\frac{8^0}{7^{-3}})^3 \times 7^{-9}$

$=\frac{8^2}{7^2}\times \frac{8^0}{7^{-3}}\times 7^{-9}$

As we know that the exponential rule :

$(\frac{a}{b})^m=\frac{a^m}{b^m}$

and

$(a^m)^n=a^{mn}$

And

$a^0=1$

By applying such rule, we get,

$\frac{8^2}{7^2}\times \frac{8^0}{7^{-3}}\times 7^{-9}\\\\=\frac{64}{49}\times \frac{1}{7^{-9}}\times 7^{-9}\\\\=\frac{64}{49}\times \frac{7^{-9}}{7^{-9}}\\\\=\frac{64}{49}$

Hence, First Option is correct.