Question

can somebody explain how to find the formula to the inverse function log2 6-log2 15+log 20

Answer 1

do you mean the inverse variation?

Answer 2

sorry find the exact value of the expression

Answer 3

do you agree that
log(x^(2)) = 2log(x)
?
and that x^(-1) = 1/x

Answer 4

yes the one to one value ?

Answer 5

log_2(6)-log_2(15)+log(20)

so log_2(6) + log_2(15^(-1)) + log(20)
=
log_2(6/15) + log20

Answer 6

you can't simplify it anymore because the the logrithims don't share the same base

Answer 7

*logarithms

Answer 8

do you want to find the inverse of this? anyfeed back would be appreciated.

BTW
you can't take log(0) because A^(x) =/= 0
A being any real number

Answer 9

what ? i thought I was suppose to divide do log2 6*20 and divide by 15...

Answer 10

thew question is

\[\log _{2} (6/15 \times20) = \log _{2}(8) = 3\]

this is becuase 2^3 = 8

Answer 11

aah yes

Answer 12

campbell_st wth?

Answer 13

repost the question so it isn't so werid

Answer 14

you can't multiply by 20 as the log is not of the same base thus they cannot be combined cambell

Answer 15

well I assumed that it was base 2.... if base 10 then use change of base on all...to say base e

Answer 16

you shouldn't assume things when it comes to this kind of math but yeah if he wrote the question with all log_2
log_2(x) = y
x = 2^(y)

then it would be
log_2((6/15)*20)
log_2(8) = y

so

8 = 2^(y)
y = 3

Answer 17

therefore,
log2 6-log2 15+log2 20 = 3