How would you write x^-2 with no negative exponents?
J JMM90 New member Joined Sep 28, 2006 Messages 9 Sep 28, 2006 #1 How would you write x^-2 with no negative exponents?
D Denis Senior Member Joined Feb 17, 2004 Messages 1,707 Sep 28, 2006 #2 Rule: a^(-x) = 1 / a^x ; ok?
skeeter Elite Member Joined Dec 15, 2005 Messages 3,215 Sep 28, 2006 #3 \(\displaystyle \L \frac{1}{x^2}\)
J JMM90 New member Joined Sep 28, 2006 Messages 9 Sep 28, 2006 #4 Thanks! Would that mean that x ^-2 + x ^3 would be 1/ x^5
skeeter Elite Member Joined Dec 15, 2005 Messages 3,215 Sep 28, 2006 #5 JMM90 said: Thanks! Would that mean that x ^-2 + x ^3 would be 1/ x^5 Click to expand... no way ... x<sup>-2</sup> + x<sup>3</sup> = 1/x<sup>2</sup> + x<sup>3</sup> = 1/x<sup>2</sup> + x<sup>5</sup>/x<sup>2</sup> = (1 + x<sup>5</sup>)/x<sup>2</sup>
JMM90 said: Thanks! Would that mean that x ^-2 + x ^3 would be 1/ x^5 Click to expand... no way ... x<sup>-2</sup> + x<sup>3</sup> = 1/x<sup>2</sup> + x<sup>3</sup> = 1/x<sup>2</sup> + x<sup>5</sup>/x<sup>2</sup> = (1 + x<sup>5</sup>)/x<sup>2</sup>
