Question

Use the various laws of logarithims to find the values of the positive integers a and b:
3log2(a )+ bLog2(4)-log2(64)=1

Answer 1

can I help

Answer 2

yes please!

Answer 3

what do you think it is

Answer 4

5(a)(i) for reference :)

Answer 5

Well so far I've simplified it to log2(a^3+4^b/64)=1, but I'm not sure what to do next

Answer 6

I've also got 2^(64-a^3) = 4^b in my working but idk where to go from there

Answer 7

great what is this from a book or flvs

Answer 8

@4meisu

Answer 9

are you there

Answer 10

` log2(a^3+4^b/64)=1`

how did you get a `+` sign in between?

Answer 11

oh yep, you're right hartnn

Answer 12

do u know that u can multiply the value when bases are same ?

Answer 13

@4meisu ?

Answer 14

and..here u have base as 2 on the LHS.. so try to write 1 as log something.base 2 OK?

Answer 15

@4meisu ? are u there?

Answer 16

so you got till here?

\(\log_2(\dfrac{a^34^b}{64})=1
\)
?

Answer 17

Yep, I got till there

Answer 18

now we can write 1 as \[\log_{2} 2\]

Answer 19

do u know that?