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
[tex](\frac{8\times 4\times 2}{8\times 7})^{2} \times (\frac{8^0}{7^{-3}})^3 \times 7^{-9}[/tex]
So, when we cancel out 8 from numerator and denominator form the first expression, we get
[tex](\frac{ 4\times 2}{7})^{2} \times (\frac{8^0}{7^{-3}})^3 \times 7^{-9}[/tex]
[tex]=\frac{8^2}{7^2}\times \frac{8^0}{7^{-3}}\times 7^{-9}[/tex]
As we know that the exponential rule :
[tex](\frac{a}{b})^m=\frac{a^m}{b^m}[/tex]
and
[tex](a^m)^n=a^{mn}[/tex]
And
[tex]a^0=1[/tex]
By applying such rule, we get,
[tex]\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}[/tex]
Hence, First Option is correct.
The value is equivalent to 64 over 49
Let's recall following formula about Exponents and Surds:
[tex]\boxed { \sqrt { x } = x ^ { \frac{1}{2} } }[/tex]
[tex]\boxed { (a ^ b) ^ c = a ^ { b . c } } [/tex]
[tex]\boxed {a ^ b \div a ^ c = a ^ { b - c } }[/tex]
[tex]\boxed {\log a + \log b = \log (a \times b) }[/tex]
[tex]\boxed {\log a - \log b = \log (a \div b) }[/tex]
Let us tackle the problem.
First, we will change this "word problem" into mathematical equation.
8 multiplied by 4 multiplied by 2 ↓
[tex]8 \times 4 \times 2 = 64[/tex]
The whole over 8 multiplied by 7 ↓
[tex]64 \div (8 \times 7) = 64 \div 56 = 8 \div 7[/tex]
The whole raised to the power of 2 ↓
[tex]( 8 \div 7 )^2[/tex]
The whole multiplied by 8 to the power of 0 over 7 to the power of negative 3, whole to the power of 3 ↓
[tex](\frac{8}{7})^2 \times (\frac{8^0}{7^{-3}})^3[/tex]
[tex](\frac{8}{7})^2 \times (\frac{1}{7^{-9}})[/tex]
[tex](\frac{8^2}{7^2}) \times (\frac{1}{7^{-9}})[/tex]
[tex]\frac{8^2}{7^{-7}}[/tex]
The whole multiplied by 7 to the power of negative 9 ↓
[tex]\frac{8^2}{7^{-7}} \times 7^{-9}[/tex]
[tex]\frac{8^2}{7^2}[/tex]
[tex]\large {\boxed {\frac{64}{49} } }[/tex]
Grade: High School
Subject: Mathematics
Chapter: Exponents and Surds
Keywords: Power , Multiplication , Division , Exponent , Surd , Negative , Postive , Value , Equivalent
