Question

Simplify each exponential expression using the properties of exponents and match it to the correct answer.

Answer 1

The match of solutions with the exponential expressions is

1) $\frac{(2(3^{-2}) )^{3} (5(3^{2}) )^{2} }{(3^{-2})((5)(2))^{2}}=2$

2) $(3^3) (4^0)^2 (3(2))^{-3} (2^2)=1/2$

3) $\frac{(3^74^7) (2(5))^{-3} (5)^2}{(12^7) (5^{-1}) (2^{-4})} =2$

4) $\frac{(2(3))^{-1} (2^0)}{(2(3))^{-1}}=1$

What are properties of exponents?

The base will be multiplied by itself a certain number of times, as indicated by the exponent (also known as a power or degree).

What are the formulae/ properties for exponents?

Formulae for solving exponents are referred to as exponents formulas. The exponent of a number is written as $x^{n}$, which means that x has been multiplied by itself n times.

$x^{n}(x^{m})=x^{n+m} \\\frac{x^{n} }{x^{m} }=x^{n-m} \\(x^{n} )^{m} =x^{nm} \\((x)(y))^{n}=x^{n}(y^{n} \\x^{0}=1$

1) the solution of the first expression will be

$\frac{(2(3^{-2}) )^{3} (5(3^{2}) )^{2} }{(3^{-2})((5)(2))^{2}}$

$(2^3) (3^{-6} ) (5^2) (3^4) / (3^{-2}) (10^2)= (2.2^2.5^2) (3^{-6}.3^4) / (3{^-2}) (10^2)= (2)(10^2) (3^{-2}) / (3^{-2}) (10^2)\\=2$

2)The solution of the second expression will be

$(3^3) (4^0)^2 (3(2))^{-3} (2^2)$

Any  number with power zero is 1.

So,

$(3^3) (1^2) (3^{-3}) (2^{-3}) (2^2)= (2^{(-3+2)})=2^-1\\= 1/2$

3) The solution of third expression will be

$\frac{(3^74^7) (2(5))^{-3} (5)^2}{(12^7) (5^{-1}) (2^{-4})} \\= \frac{(12^7) (2^{-3}) (5^{-3}) (5^2)}{(12^7) (5^{-1}) (2^{-4})} =\frac{(2^-3) (5^{(-3+2)})}{(5^{-1}) (2^{-4})} = \frac{(2^{-3}) (5^{-1})}{(5^{-1}) (2^{-}4)} = \frac{1}{(2^{-1})} = 2$

4) The solution to the forth expression will be

$\frac{(2(3))^{-1} (2^0)}{(2(3))^{-1}}\\=2^{0}\\ =1$

To learn more about properties of exponents, visit: https://brainly.com/question/17318340

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