## JessicaNicoleCraven Group Title k^3(k7/5)^-5 one year ago one year ago

1. angela210793

|dw:1352048933|dw:1352048968202:dw|095:dw|

2. CliffSedge

? Is $$\large k^3 (\frac{k^7}{5})^{-5}$$ ?

3. angela210793

:O where did my solution go :O

4. CliffSedge

^ ? I dunno, looks like you were trying to post a drawing.

5. JessicaNicoleCraven

I am so confused. Its suppose to be k with an exponent of 3, times k(7/5) with an exponent of negative 5.

6. CliffSedge

Then is it $$\large k^3 \cdot (k^{7/5})^{-5}$$ ?

7. JessicaNicoleCraven

Yeah.

8. CliffSedge

Ok. Order of operations, handle the exponent first. A power-to-a-power, multiply the exponents.

9. angela210793

All u have to know is that x^a*x^b=x^(a+b) (x^a)^b=x^ab and x^-a=1/x^a so u'll have |dw:1352049321095:dw|

10. CliffSedge

$$\huge \rightarrow k^3 \cdot k^{(\frac{7}{5} \cdot -5)}$$ $$\huge \rightarrow k^3 \cdot k^{-7}$$

11. JessicaNicoleCraven

Oh wow, thats a lot more simple than I was trying to make it. Thank you!