Indices Problem

Mc1980

Hi all,

Can anyne out there help me out with this question its riving me mad :mad:
(a^2 b)^1/3 over (a^-1 b^4)^1/3
So thats (a squared b)to the power of 1 third over a to the power of minus 1 b to the power of 4) to the power of 1 third
hope this makes sense

thanks in advance
M

sponsoredwalk

[latex] \frac{(a^2b)^{ \frac{1}{3}}}{(a^{-1}b^4)^{ \frac{1}{3}}} [/latex]

Reading these links:

http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx
http://tutorial.math.lamar.edu/Classes/Alg/RationalExponents.aspx
http://tutorial.math.lamar.edu/Classes/Alg/RealExponents.aspx

in order should teach you how to do any problem of this kind, including the
one you've got here. A little tip, if you forget the rules for exponents, as I
frequently do , just rederive them using simple examples like [latex]2^3 \cdot 2^4 \ = \ ?[/latex]
and [latex] (2^3)^2[/latex] where you can do something like
(2³)² = (8)² = 64 which means [latex] (2^3)^2 \ = \ 2^{(3 + 2)} \ = \ 2^5[/latex] has to be wrong as
[latex] 2^5 \ = \ 32[/latex] so the rule must be [latex] (2^3)^2 \ = \ 2^{(3 \cdot 2)} \ = \ 2^6 \ = \ 64[/latex]
which is correct and from that you can quickly generalize.= to rational
exponents etc...

Eliot Rosewater

sponsoredwalk wrote: »

[latex] \frac{(a^2b)^{ \frac{1}{3}}}{(a^{-1}b^4)^{ \frac{1}{3}}} [/latex]

Just a quick note, if you stick the LaTeX command \displaystyle before your LaTeX code it becomes more readable:

[latex] \displaystyle \frac{(a^2b)^{ \frac{1}{3}}}{(a^{-1}b^4)^{ \frac{1}{3}}} [/latex]