How do I write$\displaystyle (-x^y^2)^4$ without parentheses Is it $\displaystyle x^4^y^8$? explain please
Follow Math Help Forum on Facebook and Google+
Originally Posted by wolfhound How do I write$\displaystyle (-x^y^2)^4$ without parentheses $\displaystyle (-x^{y^2})^4 = (-1)^4 \cdot (x^{y^2})^4 = x^{4y^2}$
Thanks but how does the y^2 exponent stay the same?
Originally Posted by skeeter $\displaystyle (-x^{y^2})^4 = (-1)^4 \cdot (x^{y^2})^4 = x^{4y^2}$ Don't think so. We have -x^, not (-x)^ ; like, -2^2 = -4, (-2)^2 = 4
Last edited by mr fantastic; Oct 4th 2010 at 01:06 AM. Reason: Deleted unnecessary text.
Originally Posted by Wilmer Don't think so. We have -x^, not (-x)^ ; like, -2^2 = -4, (-2)^2 = 4 take another look at the original post ...
Last edited by mr fantastic; Oct 4th 2010 at 01:06 AM. Reason: Edited quote.
Hes correct, I just don't understand the way $\displaystyle y^2$ is not raised to $\displaystyle y^8$ ?
Originally Posted by wolfhound Hes correct, I just don't understand the way $\displaystyle y^2$ is not raised to $\displaystyle y^8$ ? the rule is ... $\displaystyle (x^a)^b = x^{a \cdot b}$ so, if $\displaystyle a = y^2$ and $\displaystyle b = 4$ ... what is the result?
um $\displaystyle 4y^2$ ....
Originally Posted by wolfhound um $\displaystyle 4y^2$ .... bingo
View Tag Cloud