a and b are positive integers and a-b=2. Evaluate the following:
9^1/2b/3^a

Answer:

[tex] \frac{1}{ 9 }[/tex]

Step-by-step explanation:

[tex] \huge \frac{ {9}^{ \frac{1}{2}b } }{ {3}^{a} } \\ \\ = \huge \frac{ {( {3}^{2} )}^{ \frac{1}{2}b } }{ {3}^{a} } \\ \\ = \huge \frac{ { {3}}^{2 \times \frac{1}{2}b } }{ {3}^{a} } \\ \\ \huge = \frac{ {3}^{b} }{ {3}^{a} } \\ \\ \huge = \frac{1}{ {3}^{a - b} } \\ \\ \huge= \frac{1}{ {3}^{2} } \\ ( \because \: a - b = 2)\\ \\ \huge = \frac{1}{ 9 }[/tex]

The evaluation of the expression 9^((1/2)b)/3^a is 1/9

If we have a^b then 'a' is called base and 'b' is called power or exponent and we call it "a is raised to the power b" (this statement might change from text to text slightly).

Exponentiation(the process of raising some number to some power) have some basic rules as:

[tex]a^{-b} = \dfrac{1}{a^b}\\\\a^0 = 1 (a \neq 0)\\\\a^1 = a\\\\(a^b)^c = a^{b \times c}\\\\ a^b \times a^c = a^{b+c} \\\\^n\sqrt{a} = a^{1/n} \\\\(ab)^c = a^c \times b^c\\\\a^b = a^b \implies b= c \: \text{ (if a, b and c are real numbers and } a \neq 1 \: and \: a \neq -1 )[/tex]

Now, for this case, we are given that:

a - b = 2

The expression is:

[tex]\dfrac{9^\dfrac{1}{2}b}{3^a}[/tex]

Since we know that [tex]3^2 = 9[/tex], thus:

[tex]\dfrac{9^{\frac{1}{2}b}}{3^a} = \dfrac{(3^2)^{\frac{1}{2}b}}{3^a} = \dfrac{3^{2 \times 1/2 \times b}}{3^a} = \dfrac{3^b}{3^a} = \dfrac{1}{3^{a-b}}[/tex]

Since a-b = 2, therefore, we get:
[tex]\dfrac{9^{\frac{1}{2}b}}{3^a} =\dfrac{1}{3^{a-b}} = \dfrac{1}{3^2}} = \dfrac{1}{9}[/tex]

Thus, the evaluation of the expression 9^((1/2)b)/3^a is 1/9

Learn more about exponentiation here:

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