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If a = -1 and b = 2, find :

(i) ab+baa^b + b^a

(ii) ab−baa^b - b^a

(iii) ab×baa^b \times b^a

(iv) ab÷baa^b \div b^a

Indices

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Answer

(i) ab+baa^b + b^a

Substituting the values a = -1 and b = 2,

=(−1)2+2−1=1+12=2+12=32=112= (-1)^2 + 2^{-1}\\[1em] = 1 + \dfrac{1}{2}\\[1em] = \dfrac{2 + 1}{2}\\[1em] = \dfrac{3}{2}\\[1em] = 1\dfrac{1}{2}

Hence, ab+ba=112a^b + b^a = 1\dfrac{1}{2}.

(ii) ab−baa^b - b^a

Substituting the values a = -1 and b = 2,

=(−1)2−2−1=1−12=2−12=12= (-1)^2 - 2^{-1}\\[1em] = 1 - \dfrac{1}{2}\\[1em] = \dfrac{2 - 1}{2}\\[1em] = \dfrac{1}{2}

Hence, ab−ba=12a^b - b^a = \dfrac{1}{2}.

(iii) ab×baa^b \times b^a

Substituting the values a = -1 and b = 2,

=(−1)2×2−1=1×12=12= (-1)^2 \times 2^{-1}\\[1em] = 1 \times \dfrac{1}{2}\\[1em] = \dfrac{1}{2}

Hence, ab×ba=12a^b \times b^a = \dfrac{1}{2}.

(iv) ab÷baa^b \div b^a

Substituting the values a = -1 and b = 2,

=(−1)2÷2−1=1÷12=2= (-1)^2 \div 2^{-1}\\[1em] = 1 \div \dfrac{1}{2}\\[1em] = 2

Hence, ab÷ba=2a^b \div b^a = 2.

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