Question

Neeed help..!!

Answer 1

fill in the blank
\[2^{blank}=64\]

Answer 2

6

Answer 3

got the first one!

Answer 4

\[64^{blank}=2\] this one is trickier

Answer 5

yeaa... ummmm

Answer 6

hint, how do you write "sixth root" as an exponent?
that is how you you write
\[\sqrt[6]{64}=2\] using exponential notation?

Answer 7

1/6

Answer 8

ok two down!

Answer 9

:]

Answer 10

\[81^{blank}=\frac{1}{3}\] another tricky one

Answer 11

hint: first take the fourth root, then take the reciprocal

Answer 12

ok

Answer 13

81^1/4

Answer 14

3=1/3

Answer 15

that will get you from 81 to 3
to get to \(\frac{1}{3}\) write
\[81^{-\frac{1}{4}}=\frac{1}{3}\]so answer to third one is
\[\log_{81}(\frac{1}{3})=-\frac{1}{4}\]

Answer 16

then negative exponent means take the reciprocal

Answer 17

ohh okay got it!!

Answer 18

good because you are going to need it for the next one too!

Answer 19

yea im trying!

Answer 20

\[(\frac{1}{3})^{blank}=27\]

Answer 21

flip it, then cube it

Answer 22

-3

Answer 23

ok! got it! one more to go

Answer 24

its hard..!

Answer 25

maybe better to rewrite as
\[\log_2(\sqrt{2})-\log_2(4)\] the second one is easy

Answer 26

ohh yea division rule..

Answer 27

yea its 2

Answer 28

ohh the first one is 2^(1/2)

Answer 29

so 2^(1/2)-2

Answer 30

last one??

Answer 31

This is a perfect example of a tutoring!! All hail @satellite73 !!

Answer 32

Agree @apoorvk. Great work @satellite73 !!