A-Level Edexcel Maths Pure Radian Measure: 2. (a) Write down the value of $

2. (a) Write down the value of $32^{rac{1}{3}}$ - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 1

Question 5

$2.-(a)-Write-down-the-value-of-$32^{-rac{1}{3}}$-Edexcel-A-Level Maths Pure-Question 5-2014-Paper 1.png$

2. (a) Write down the value of $32^{rac{1}{3}}$. (b) Simplify fully $(32x^{-5})^{-rac{3}{2}}$.

Worked Solution & Example Answer:2. (a) Write down the value of $32^{rac{1}{3}}$ - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 1

Step 1

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Answer

To find the value of $32^{ rac{1}{3}}$ , we can recognize that $32$ is equal to $2^5$ . Therefore:

$32^{ rac{1}{3}} = (2^5)^{ rac{1}{3}} = 2^{ rac{5}{3}}.$

Thus, the final answer is:

$2^{ rac{5}{3}}$ .

Step 2

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Answer

Let's break down the expression step by step:

Start by rewriting $32$ as $2^5$ :

$(32x^{-5})^{- rac{3}{2}} = (2^5 x^{-5})^{- rac{3}{2}}.$
Using the property of exponents, we can distribute the exponent:

$= (2^5)^{- rac{3}{2}} (x^{-5})^{- rac{3}{2}}.$
Simplifying each part:
- For the first part: $= 2^{5 imes - rac{3}{2}} = 2^{- rac{15}{2}}.$
- For the second part:
  $= x^{-5 imes - rac{3}{2}} = x^{ rac{15}{2}}.$
Combining these results gives:

$= 2^{- rac{15}{2}} x^{ rac{15}{2}}.$
Finally, writing the negative exponent in the denominator results in:

$= rac{x^{ rac{15}{2}}}{2^{ rac{15}{2}}} = rac{x^{ rac{15}{2}}}{ ext{(sqrt of } 2^{15})} = rac{x^{ rac{15}{2}}}{4 ext{ (as } 2^{ rac{15}{2}} = 4 ext{)}}.$