Question

I have 2 questions. Can someone help me?

Answer 1

Sure

Answer 2

Qustion 1

What value of k solves the equation?

-64

- 4

4

64

Answer 3

*question

Answer 4

Alright so you have to find k :P

k^-4 signifies 1/(k^4) so you are actually looking for k^4=256 so the bottoms of the fractions are the same

Answer 5

And the fourth root of 256 is 4 (ie) 4*4*4*4 or 16*16 = 256

Answer 6

So your answer will be (c) 4

Answer 7

Ok thanks so much!

Answer 8

Did you have another question?

Answer 9

yes let me upload it ☺

Answer 10

Or you can write it in, either way, and I will try to assist you

Answer 11

Ok thanks!!!

Answer 12

What value of n solves the equation?

4

3

3

4

Answer 13

I think your post removed your negative signs in the answers :P

Answer 14

Oh yes it did. Sorry lemme re write it

Answer 15

An easy way to think about this one is to use the same rule we used in the last problem, where 2^(-n)=8

Answer 16

Its alright don't bother :)

Answer 17

For this, you can simply take the log(base2) of both sides, yielding you -n=log(base2) of 8

which you can change to -n=log(base2) of (2^3)

Answer 18

Ok thanks so muvh for helping!! I really apperciate!!!!

Answer 19

*much

Answer 20

I am assuming you have learned that the log(base a) of a^b is just b correct? as in the log of 10^2 is 2

Answer 21

if not I can show you a different way

Answer 22

No, I don't recall that

Answer 23

Alright here I can do this instead

Answer 24

So 2^n=1/8... you can rewrite this as 2^n=2^(-3) because 2^(-3) is like saying 1/(2*2*2) which is 1/8... so there you have the same base and different exponents so you can set the exponent parts equal, so n=-3

Answer 25

and your answer should be (b) or (c), whichever is -3

Answer 26

Ok thanks so much!!! You have big a HUGE help. I understand it now.