# Fractional Indices

• Feb 24th 2009, 05:41 AM
sammy28
Fractional Indices
$4n^{3/2} = 8^{-1/3}$

Find the value of n

hi all, ive been given the above and need to solve it without a calculator. ive expessed it in all the indice rules but cant come up with a suitable format to solve it.

any ideas?

here's what i got so far $n^{3/2} = \frac{1}{8}$
so does $n = \sqrt[{3/2}]{1/8}$

$(n^{3/2})^{2/3}=(2^{-3})^{2/3}$
therefore
$n=2^{-2}$
therefore
$n=\frac{1}{2^2}$
• Feb 24th 2009, 06:43 AM
(Hi)Welcome to the Forum(Party)

$4 = 2^2$

$8 = 2^3$

using it in
$4n^{3/2} = 8^{-1/3}$

we get
$2^2 n^{3/2} = (2^{3})^{-1/3}$

divide both sides by 2^2

$n^{3/2} = \frac {2^{-1} }{2^2}$

Thus
$n^{3/2} = 2^{-1 - 2}$

Hence

$n^{3/2} = 2^{-3}$

This can be written as

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

Hence

$
n^1 = \frac{1}{2^{(2*3)/3}}
$

Hence $n= \frac{1}{4}$
• Feb 24th 2009, 06:49 AM
sammy28
wow thanks for taking the time to look at that one. i thought id covered them all except i didnt think about converting 2 to the power of n as a first step. and multiplying by the reciprocal.

now i can do it without a calculator (Clapping)

thanks for the welcome too. how can you format questions like your answer, ie using mathematical symbols?
• Feb 24th 2009, 09:08 AM