Indices Question grade A*

Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

Answer 2

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 $2/3$: $$(n^{3/2})^{2/3}=\left(\frac{1}{16}\right)^{2/3}$$ $$n^{\frac{3}{2}\cdot \frac{2}{3}}=\left(\frac{1}{16}\right)^{2/3}$$ $$n=\left(\frac{1}{16}\right)^{2/3}$$ $$n=\frac{1}{16^{2/3}}=\frac{1}{2^{8/3}}=\frac{1}{4\times 2^{2/3}}$$