Question

How do i do this?????

Answer 1

by remembering what the properties of exponents and logs are :)

Answer 2

\[b^{n+m}=b^n*b^m\]

Answer 3

i know that 4^-3 x 4^2log2(7)

Answer 4

good so far

Answer 5

:]

Answer 6

now is that 4^-3 x 2log2(7)=4?

Answer 7

rewrite 2log2(7) as log2(49)

Answer 8

okay

Answer 9

if you introduce an = sign; you have to put an arbitrary variable over the to equate to

Answer 10

okay but is it okay to write like that?

Answer 11

\[4^{log_2(49)}=4k\]is better

Answer 12

okay \[\log _{4}81\] do u know how to write this function in the calculator?

Answer 13

lost a ^3

Answer 14

\[4^{-3}\times2\log _{2}49\]
0.015625x3.38
0.0528 u mean like this?

Answer 15

\[4^{-3}4^{log_2{49}}=k\]

\[4^{log_2{49}}=4^3k\]

\[log_4\{4^{log_2{49}}=4^3k\}\]

\[log_2(49)=3+log_4(k)\]

\[\frac{ln(49)}{ln(2)}=3+\frac{ln(k)}{ln(4)}\]

Answer 16

k can be any variable like x?

Answer 17

\[\frac{ln(4)ln(49)}{ln(2)}-3ln(4)=ln(k)\]

hmmm, we can e both sides but that doesnt seem like a simplification does it

Answer 18

yes, its just a filler that you try to solve for

Answer 19

i use k for some constant

Answer 20

okay thanks alot :]

Answer 21

play around with it using the basic rules you know for exponents and logs and you should do fine

Answer 22

then use wolfram for a dbl chk ;)

Answer 23

3.621117=ln k