Question

;)))

Answer 1

Similar Problem:


\[\Large \log_{3}(60) - \log_{3}(12)\]


\[\Large \log_{3}\left(\frac{60}{12}\right)\]


\[\Large \log_{3}(5)\]


So \[\Large \log_{3}(60) - \log_{3}(12) = \log_{3}(5)\]

Answer 2

that's all i have to do?

Answer 3

pretty much, but just use your numbers in place of mine

Answer 4

yeah, thanks! do you know how to do the other one?

Answer 5

in general

\[\Large \log_{b}(x) - \log_{b}(y) = \log_{b}\left(\frac{x}{y}\right)\]

Answer 6

what do 16 and 64 have in common

Answer 7

they both have a 6 in them, both even #'s

Answer 8

good, but in terms of factors

Answer 9

2^6? i don't know :S :/

Answer 10

both are powers of 2 or powers of 4

Answer 11

ahh, yeah

Answer 12

so

16 = 2^4
64 = 2^6

and

16 = 4^1
64 = 4^3

Answer 13

that's it?

Answer 14

well you need to rewrite both sides to have the same base

Answer 15

for instance, say you had

2^x = 16

you would rewrite 16 as 2^4 to get

2^x = 2^4

and you can see that x must be 4 (since both bases are 2)

Answer 16

so you're using this idea to solve the second problem

Answer 17

ohhh, well that's pretty simple! thank youu

Answer 18

actually waait, it's not! haha.. i just got confused

Answer 19

with what

Answer 20

where are you stuck

Answer 21

and you can see that x must be 4 (since both bases are 2) ??

Answer 22

well if b^x = b^y, then x = y

Answer 23

i'm confused because you did 2 different problems..:/

Answer 24

that's just an example

Answer 25

okay, that makes sense .. but i'm still confused on what to do after what you said ''16 = 2^4
64 = 2^6

and

16 = 4^1
64 = 4^3''

Answer 26

oh i meant to say 16 = 4^2, my bad

Answer 27

1/16=64^(4x-3)

1/(4^2) = (4^3)^(4x-3) ... since 16 = 4^2 and 64 = 4^3

4^(-2) = 4^(3*(4x-3))

so because the bases are now the same, we can say that the exponents must be the same

so...

-2 = 3*(4x-3)

solve for x

Answer 28

ahh, got it :) thanks a lot!!!

Answer 29

np