## anonymous one year ago What is the value of the expression? A. 9−48 B. 9−2 C. 92 D. 922

1. anonymous

2. anonymous

Do you remember your exponent rules?$a^b\times a^c = a^{b + c}$$\frac{a^b}{a^c}=a^{b-c}$So, try working this one step at a time.$\frac{9^8}{9^{-4}\times9^{10}}=\frac{9^8}{9^{-4}}\times\frac{1}{9^{10}}$Does this help?

3. anonymous

Confusing... @Nnesha

4. anonymous

What part is confusing you?

5. anonymous

every part...

6. Nnesha

1st write those exponents rules on a piece of paper(or notebook)!!!!!!!!PLEASE!!!

7. anonymous

ok

8. anonymous

done

9. Nnesha

no.

10. anonymous

huh i meant done as in im done writing them down

11. Nnesha

did you write all of them just in 2 minutes ?

12. anonymous

yes i did ill show u

13. Nnesha

14. anonymous

15. anonymous

16. Nnesha

looks good.

17. anonymous

umm i guess she meant u had to help me @Calcmathlete

18. anonymous

*thinking* -I'm to stupid for her-

19. anonymous

Alright. So, the first one. ($$a^b \times a^c = a^{b+c}$$). Let's do a smaller example so you can see what this means. $2^3\times 2^2 = ?$These numbers are small enough that we can do it by hand. $$2^3 = 8$$ and $$2^2 = 4$$, right? If we use this method, $$2^3\times2^2 = (8)\times(4) = 32$$. However, if you notice, if we add the exponents instead, $$2^3 \times 2^2 = 2^{3 + 2} = 2^5 = 32$$. Do you see what happened there?

20. anonymous

yes

21. anonymous

u did 2*2*2 which = 8 and then 2*2 which = 4 then u multiplied them and got 32

22. anonymous

And if you're wondering why this works, think about it this way. When you have $$2^3$$, what you're really doing is multiplying 2 by itself 3 times. So, $$2^3 = 2\times2\times2$$. So, it should also make sense that $$2^3 \times2^2 = (2\times2\times2)\times(2\times2) = 2\times2\times2\times2\times2 = 2^5$$ Does this part make sense?

23. anonymous

yes

24. anonymous

Ok. Let's work on the original problem then. Work on it little by little. The entire thing is $\frac{9^8}{9^{-4}\times 9^{10}}$So, let's work with the denominator first. Using what we did above (adding exponents), $$\large{9^{-4} \times9^{10}=9^{?}}$$

25. anonymous

so would u subtract the exponents now or just multiply to get -40

26. anonymous

Well, adding a negative number is subtracting, technically. If we follow what we did above, you would add (-4) to 10.

27. anonymous

so 6

28. anonymous

Yes. So, we can say that $$9^{-4}\times9^{10} = 9^6$$. From there, plug it back in and move onto the second step. $\frac{9^8}{9^{-4}\times9^{10}} \rightarrow\frac{9^8}{9^6}$Here, we're dividing instead of multiplying. So, what would you do with the exponents now if we follow the fact that $$\frac{a^b}{a^c}=a^{b-c}$$?

29. anonymous

In other words, $$\large{\frac{9^8}{9^6}=9^{?}}$$

30. anonymous

umm do i just use the exponents like 8-6 or something

31. anonymous

cuz thats 2 so that means its c

32. anonymous

Yes. You're dividing, so you subtract the exponents. And yes, it is C. Good job.

33. anonymous

Yay tysm!!!

34. anonymous

I have 1 more question could u help me with it in a new post?

35. anonymous

Sure.