Question

Using log8 10=P , log8 12=Q, log8 11=R Find log8 5/256

Answer 1

do you mean express it in letters?

Answer 2

start with
\[\log_8(5)-\log_8(256)\]

Answer 3

Yes >.<

Answer 4

then we see if we can write 5 and 256 with the numbers given.

Answer 5

i think we're gonna need to add a few lol

Answer 6

damn i got stuck a second hold on
we know
\[256=2^8\] so using base 8 we can write it as
\[8^{\frac{8}{3}}\] so we know
\[\log_8(256)=\frac{8}{3}\]

Answer 7

Its hard huh ? I know the answer but i want to know how to get there.

Answer 8

and since \(5=\frac{10}{2}\) we have
\[\log_8(10)=\log_8(5)-\log_8(2)\]

Answer 9

\[\log_8 5 = \log_8 10 - \log_8 2\]

Answer 10

\[\log_8(2)=\frac{1}{3}\] because
\[8^{\frac{1}{3}}=2\]

Answer 11

It's P - 3 , some how.

Answer 12

and you are given that
\[\log_8(5)=P\]so your final answer is
\[P-\frac{1}{3}-\frac{8}{3}=P-3\]

Answer 13

i sense that this was not clear, so should we run through the steps again?

Answer 14

Answer is P-3 . but P = log8 10

Answer 15

O.O

Answer 16

right so lets make sure it is clear what happened ok?

Answer 17

okay (:

Answer 18

\[\log_8(5)=\log_8(\frac{10}{2})=\log_8(10)-\log_8(2)\]

Answer 19

why did we write it this way? because you were given that
\[\log_8(10)=P\] so we could use that

Answer 20

then since we can write
\[8^{\frac{1}{3}}=2\] we know
\[\log_8(2)=\frac{1}{3}\]

Answer 21

And your using one of the properties of logarithms in which log x/y=log x-log y

Answer 22

therefore we know
\[\log_8(5)=P-\frac{1}{3}\]
yes that is what i am using

Answer 23

i also used it at the bigginning when i wrote
\[\log_8(\frac{5}{256})=\log_8(5)-\log_8(256)\]

Answer 24

@satellite73 got drunk a while ago and wrote \[\large \log_8 10 = \log_8 5 - \log_8 2\] lol

Answer 25

yes you are right (on both counts)
where did you get the bike??

Answer 26

idk...i just googled red bike haha =))

Answer 27

Try this one :)
\[\log _{7}3 = X , \log _{7}8 = Y , \log _{7}10=Z
Find \log _{7} 15/32\]