Easy Indices

By rule of adding and subtracting powers : $a^m \times a^n = a^{m+n}$ and $a^m \div a^n = a^{m-n}$ .

For the numerator, we have $2^5 \times 2^7 = 2^{5+ 7} = 2^{12}$ .

For the denominator, we have $2^{10} \times 2^3 = 2^{10+3} = 2^{13}$ .

So the $\frac{2^5 \times 2^7 }{2^{10}\times2^3} = \frac{2^{12}}{2^{13}} = 2^{12 - 13} = 2^{-1}$ .

Because $a^{-m} = \frac1{a^m}$ and $a^0 = 1$ , we have $2^{-1} \times3^0 = \frac12 \times 1 = \frac12$ .

You need to know that $x^{n} \times x^{m}=x^{n + m}$ $\frac{x^{n}}{x^{m}}=x^{n - m}$ $x^{-n}=\frac{1}{x^{n}}$ $x^{0}=1$ - this is true for any number, and pretty logical too - if 4 is 2x2 or 8/2 or 2^2, then 2 is 2x1 or 4/2 or 2^1, and 1 is 2x0.5 or 1/2 or 2^0

So, $2^{5} \times 2^{7}=2^{5 + 7}=2^{12}$

And $2^{10} \times 2^{3}=2^{10 + 3}=2^{13}$

Then $\frac{2^{12}}{2^{13}}=2^{12 - 13}=2^{-1}$

$2^{-1}=\frac{1}{2^{1}}=\frac{1}{2}$

Since $3^{0}=1$ it just stays as $\frac{1}{2}$

Let us first simplify ... $\frac { 2^{ 5 }*2^{ 7 } }{ 2^{ 10 }*2^{ 3 } }$

using the Rule of Multiplying Exponents , as follows : $\frac { 2^{ 5+7 } }{ 2^{ 10+3 } }$ , which simplifies to: $\frac { 2^{ 12 } }{ 2^{ 13 } }$

By the rule of dividing exponents, we must $2^{ 12-13 }$ , which we can simplify to the negative exponent: $2^{ -1 }$ , which becomes ${ \frac { 1 }{ 2 } }$ .

Final Answer : ${ \frac { 1 }{ 2 } }$

$Q\bullet E\bullet D$

$\frac{2^5 \times 2^7}{2^{10} \times 2^3} \times 3^0=\frac{2^{5+7}}{2^{10+3}} \times 1=\frac{2^{12}}{2^{13}}=2^{12-13}=2^{-1}=\frac{1}{2}$

2^5x2^7x3^0/2^10x2^3

=2^12/2^13

=2^-1

=1/2. [ anything ^o=1]

3^0 is 1. We can erase that. The numerator result is 2^12 and the denominator result is 2^13. Since 2^12 is 1/2 of 2^13, the answer is 1/2

denominator is 2^12, numerator is 2^13, therefore giving 1/2

$\frac{2^5 \times 2^7}{2^{10} \times 2^3} \times 3^0$

Because $a^0 = 1$ , we have:

$\frac{2^{5+7}}{2^{10+3}} \times 1$

$= \frac{2^{12}}{2^{13}} \times 1$

$= \frac{2^0}{2^1} \times 1$

$=\frac{1}{2} \times 1$

$= \frac{1}{2}$