Question

If \[a = 3\] and \[b = - 2\], find the value of \[{a^a} + {b^b}\].

Answer 1

Hint: Here, we need to find the value of the expression \[{a^a} + {b^b}\]. We will use the given information and the rules of exponents to simplify the value of the expression \[{a^a} + {b^b}\] and obtain the required answer.

Formula Used:
The number \[{a^{ - b}}\] is equal to the reciprocal of the number \[a\] raised to the positive power \[b\], that is \[{a^{ - b}} = {\left( {\dfrac{1}{a}} \right)^b}\].

Complete step-by-step answer:
We need to find and simplify the value of the expression \[{a^a} + {b^b}\].
Substituting \[a = 3\] and \[b = - 2\] in the expression \[{a^a} + {b^b}\], we get
\[ \Rightarrow {a^a} + {b^b} = {3^3} + {\left( { - 2} \right)^{ - 2}}\]
We know that a number \[x\] raised to the exponent \[y\] can be written as \[{x^y} = x \times x \times x \times \ldots \ldots \ldots {\rm{ }}y{\rm{ times}}\].
Therefore, we can rewrite \[{3^3}\] as \[3 \times 3 \times 3\].
Thus, the expression becomes
\[ \Rightarrow {a^a} + {b^b} = 3 \times 3 \times 3 + {\left( { - 2} \right)^{ - 2}}\]
Now, we will rewrite the negative exponent as a positive exponent.
Suppose that a number \[x\] is raised to the negative power \[ - y\].
Then, \[{x^{ - y}}\] is equal to the reciprocal of the number \[x\] raised to the positive power \[y\], that is \[{x^{ - y}} = {\left( {\dfrac{1}{x}} \right)^y}\].
Substituting \[x = - 2\] and \[y = - 2\] in the equation, we get
\[ \Rightarrow {\left( { - 2} \right)^{ - 2}} = {\left( {\dfrac{1}{{ - 2}}} \right)^2}\]
The number \[{\left( {\dfrac{a}{b}} \right)^m}\] can be written as \[{\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}\].
Thus, we get
\[ \Rightarrow {\left( { - 2} \right)^{ - 2}} = \dfrac{{{1^2}}}{{{{\left( { - 2} \right)}^2}}}\]
Simplifying the expression, we get
\[ \Rightarrow {\left( { - 2} \right)^{ - 2}} = \dfrac{{1 \times 1}}{{\left( { - 2} \right) \times \left( { - 2} \right)}}\]
Substituting \[{\left( { - 2} \right)^{ - 2}} = \dfrac{{1 \times 1}}{{\left( { - 2} \right) \times \left( { - 2} \right)}}\] in the equation \[{a^a} + {b^b} = 3 \times 3 \times 3 + {\left( { - 2} \right)^{ - 2}}\], we get
\[ \Rightarrow {a^a} + {b^b} = 3 \times 3 \times 3 + \dfrac{{1 \times 1}}{{\left( { - 2} \right) \times \left( { - 2} \right)}}\]
The number \[\left( { - 2} \right)\] can be written as the product of the positive integer 2 and the negative integer \[\left( { - 1} \right)\].
Thus, we get
\[ \Rightarrow {a^a} + {b^b} = 3 \times 3 \times 3 + \dfrac{{1 \times 1}}{{\left( { - 1} \right) \times \left( { - 1} \right) \times 2 \times 2}}\]
The product of \[\left( { - 1} \right)\] and \[\left( { - 1} \right)\] is 1.
Multiplying the terms of the expression, we get
\[\begin{array}{l} \Rightarrow {a^a} + {b^b} = 27 + \dfrac{1}{{1 \times 4}}\\ \Rightarrow {a^a} + {b^b} = 27 + \dfrac{1}{4}\end{array}\]
Rewriting the expression, we get
\[ \Rightarrow {a^a} + {b^b} = \dfrac{{27}}{1} + \dfrac{1}{4}\]
The L.C.M. of 1 and 4 is 4.
Rewriting the term \[\dfrac{{27}}{1}\] with the denominator 4 by multiplying and dividing by 4, we get
\[ \Rightarrow {a^a} + {b^b} = \dfrac{{108}}{4} + \dfrac{1}{4}\]
Adding the terms in the expression, we get
\[\begin{array}{l} \Rightarrow {a^a} + {b^b} = \dfrac{{108 + 1}}{4}\\ \Rightarrow {a^a} + {b^b} = \dfrac{{109}}{4}\end{array}\]
\[\therefore \] We get the value of the expression \[{a^a} + {b^b}\] as \[\dfrac{{109}}{4}\].

Note: The terms in the given expression \[{a^a} + {b^b}\] are in exponential form. The exponential form of a number is written as \[{x^y}\], where \[x\] is called the base and \[y\] is called the exponent. \[{x^y}\] means \[x\] raised to the power of \[y\], that is \[x \times x \times x \times \ldots \ldots \ldots \] upto \[y\] times.