Question

If log base b of 2 =x and log base b of 3 =y, evaluate the following in terms of x and y:
a). log base b of 648
b). log base b of (8/81)
c). log base b of 81/log base b of 4

Answer 1

HI!!

Answer 2

we have to write 348 in term of 2 and 3

Answer 3

by the miracle of the wolf, it turns out that
\[648=2^3\times 3^4\]
http://www.wolframalpha.com/input/?i=648

Answer 4

that means
\[\log_b(648)=\log_b(2^3\times 3^4)\] now we use two properties of the log

Answer 5

so would the answer be x^3 times y^4?

Answer 6

oh no lets go slow

Answer 7

brb

Answer 8

\[\log(AB)=\log(A)+\log(B)\] so the first step it two write
\[\log_b(2^3\times 3^4)=\log(2^2)+\log(3^4)\]

Answer 9

ok i will wait

Answer 10

typo there anyways, it should be \[\log_b(2^3\times 3^4)=\log(2^3)+\log(3^4)\]

Answer 11

@xapproachesinfinity

Answer 12

hmm what the problem, didn't misty help you?

Answer 13

she did but then i left so she didnt finish helping me

Answer 14

the key is the write the number as power with the base 2 and 3

Answer 15

picking from where she left you need to use the power property of logs

Answer 16

to bring down the power

Answer 17

3log2 + 4log3

Answer 18

yes

Answer 19

and log2=x
log3=y
so it would be 3x+4y

Answer 20

ok i get it now thnx

Answer 21

no problem
it is the same thing with the other one
write them into powers then use the same method

Answer 22

ok thank you

Answer 23

i guess woflram would help if you are lazy hehe :)

Answer 24

aha yass xp

Answer 25

oh the second is letter different
first you use anther property of log \[\log(\frac{a}{b})=\log a- \log b\]

Answer 26

so =log8-log81
=log 2^3 -log 3^4

Answer 27

yeah i was doing that

Answer 28

:)

Answer 29

can you help with the last one?

Answer 30

eh that should be easy log 3^4 / log 2^2

Answer 31

ok i got 4y/2x