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I am a student and I am having difficulty with answering this question. I keep getting the answer wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.

$$(4n)^{3/2} = 8^{-1/3}$$

Find the value of $n$.

This is what I did:

$4n^{3/2} = \frac12$

I don'the know what to do next.

Thank you

callculus42
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OliviaAages
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  • Take the square on both sides of the equation. $16n^3=1/4\Rightarrow 4^2n^3=1/4\Rightarrow n^3=\frac{1}{4^3}$ – callculus42 Jan 29 '17 at 07:01
  • How do you work out what n is – OliviaAages Jan 29 '17 at 07:08
  • Take the third root on both sides. The equation can be also written as $n^3=\left(\frac{1}{4}\right)^3$ because of $\left(\frac{1}{4}\right)^3=\frac{1^3}{4^3}=\frac{1}{4^3}$ – callculus42 Jan 29 '17 at 07:12
  • How do you take the root on both sides? – OliviaAages Jan 29 '17 at 07:14
  • Calculating the root is the opposite of calculating the power. By taking the third root you neutralize the powers. $\sqrt[3]{n^3}=\sqrt[3]{\frac{1}{4^3}}$ What do you get ? – callculus42 Jan 29 '17 at 07:20
  • 1/4 is the answer to the cube root of 1/4^3 – OliviaAages Jan 29 '17 at 07:24
  • I agree to your solution :) – callculus42 Jan 29 '17 at 07:27
  • I think it is not necessary. Firstly a step by step solution has been already posted. And secondly if someone interested in the steps we (mostly I) made she/he can read the comments. But it is a good attitude that you think of other users. – callculus42 Jan 29 '17 at 07:35
  • @OliviaAage: what you did is not correct. The solutions which have been posted are solutions to what you did,but not go your original problem, since filling in 1/4 in your original problem would give 1 = 1/2 which is clearly not correct – Student Jan 29 '17 at 07:44

2 Answers2

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$(4n)^{\frac 32} = 8^{\frac {-1}3}$

$4^\frac 32.n^{\frac 32} = (2^3)^{\frac {-1}3}$

$8.n^{\frac 32} = 2^{-1}$

$8n^{\frac 32} = \frac 12$

$n^{\frac 32} = \frac 1{16}$

$n^{\frac 32} = \frac 1{2^4}$

$n^{\frac 32} = 2^{-4}$

$n = 2^{-4.\frac 23}$

$n = 2^{\frac{-8}{3}}$

$n = \frac{1}{2^{\frac{8}{3}}}$

Kanwaljit Singh
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We have: $$(4n)^{3/2}=8^{-1/3}$$ Note that the left hand side of your first step is wrong, the $4$ must also be distributed. $$4^{3/2}\cdot n^{3/2}=\frac{1}{2}$$ Giving: $$8\cdot n^{3/2}=\frac{1}{2}$$ Dividing both sides by $8$: $$n^{3/2}=\frac{1}{16}$$ Exponentiating both sides by $2/3$: $$(n^{3/2})^{2/3}=\left(\frac{1}{16}\right)^{2/3}$$ $$n^{\frac{3}{2}\cdot \frac{2}{3}}=\left(\frac{1}{16}\right)^{2/3}$$ $$n=\left(\frac{1}{16}\right)^{2/3}$$ $$n=\frac{1}{16^{2/3}}=\frac{1}{2^{8/3}}=\frac{1}{4\times 2^{2/3}}$$

projectilemotion
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