## calculusxy one year ago Help with exponents! Question posted below... MEDAL!

1. calculusxy

$\huge (x^4)^{-3} \times 2x^4$

2. calculusxy

@hartnn

3. hartnn

first term $$\huge (a^b)^c = a^{bc}$$

4. hartnn

you subtracted the exponents, the rule tell us to multiply them :3

5. anonymous

@hartnn agree

6. calculusxy

Oh my godd

7. calculusxy

I am running our of my mind :\

8. calculusxy

*out

9. hartnn

$$4 \times (-3) = - (4\times 3) = ..$$

10. calculusxy

$\huge x^{4 \times (-3)} = x^{-12}$

11. hartnn

yussss

12. hartnn

now $$\huge a^b \times a ^c = a^{b+c}$$

13. hartnn

2 $$(x^{-12} \times x^4) = .. ?$$

14. calculusxy

So here's the thing, I don't why we have to combine them... Is it because 2x^4 has two terms: 2 and x^4?

15. Jhannybean

theres a common base, $$x$$

16. hartnn

2 is a constant, lets throw that out of the process of multiplying 'x's

17. calculusxy

So just to clarify, if we have like 45x^5 then i will just have to take out the 45 and work with the x being the common base with another power right?

18. hartnn

....with another term of x. yes.

19. calculusxy

$\large 2(x^{-8}) = 2x^{-8}$

20. calculusxy

Is that correct ?

21. hartnn

$$2 x^3 \times 45 x^5 = 2\times 45 x^{3+5}$$

22. hartnn

yes, thats correct :)

23. hartnn

2/x^8 also "looks" nice

24. calculusxy

My answer is $\huge \frac{ 2 }{ x^8 }$

25. hartnn

$$\huge \checkmark$$

26. calculusxy

The reason for why I have to put the x^8 to the denominator is because of the exponent of -8 right? The negative exponent tells the base to get to its reciprocal and then the exponent becomes positive?

27. hartnn

correcto! $$\Large a^{-b} = \dfrac{1}{a^b}$$

28. calculusxy

Thank you!

29. hartnn

welcome ^_^