A-Level Edexcel Maths Pure Differentiation: 2. (a) Evaluate 81^{rac{3}{12}}

2. (a) Evaluate 81^{rac{3}{12}} (b) Simplify fully x^{2}igg(4x^{-rac{1}{2}}igg)^{2} - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 2

Question 4

$2.-(a)-Evaluate-81^{-rac{3}{12}}--(b)-Simplify-fully----x^{2}igg(4x^{--rac{1}{2}}igg)^{2}-Edexcel-A-Level Maths Pure-Question 4-2014-Paper 2.png$

2. (a) Evaluate 81^{rac{3}{12}} (b) Simplify fully x^{2}igg(4x^{-rac{1}{2}}igg)^{2}

Worked Solution & Example Answer:2. (a) Evaluate 81^{rac{3}{12}} (b) Simplify fully x^{2}igg(4x^{-rac{1}{2}}igg)^{2} - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 2

Step 1

96%

114 rated

Answer

To evaluate $81^{ rac{3}{12}}$ , we can start by simplifying the exponent. Notice that $rac{3}{12}$ can be reduced to $rac{1}{4}$ . Thus, we have:

81^{ rac{3}{12}} &= 81^{ rac{1}{4}} \ &= (3^4)^{ rac{1}{4}} \ &= 3^{4 imes rac{1}{4}} \ &= 3^{1} = 3. during the calculation, we can also say that $81 = 3^4$ to arrive at the same result.$$ We can also write: $$81^{ rac{3}{12}} = (81^{ rac{1}{2}})^{ rac{3}{6}} = (9)^{ rac{3}{6}} = 9^{ rac{1}{2}} = 3$$ Thus, the evaluation of $81^{ rac{3}{12}}$ gives us $3$.

Step 2

99%

104 rated

Answer

To simplify the expression $x^{2}(4x^{- rac{1}{2}})^{2}$ , we can start by simplifying the term $(4x^{- rac{1}{2}})^{2}$ :

(4x^{- rac{1}{2}})^{2} &= 4^{2} (x^{- rac{1}{2}})^{2} \ &= 16x^{-1}. defined Next, we can then substitute back into the expression: $$x^{2}(4x^{- rac{1}{2}})^{2} = x^{2}(16x^{-1}) = 16 x^{2 - 1} = 16x. d finally, we arrive at the fully simplified form: $16x$.