Question

Sove for x:
\[\log_2({8x})-\log_2({x^2-1})=\log_2{3}\]

Answer 1

combine the fractions and make both sides exponents to the base 2

Answer 2

i mean combine the logarithms

Answer 3

remember

log(x) - log(y) is the same as log(x/y)

Answer 4

and also remember that

\[2^{\log_2(x)} = x\]

Answer 5

when you combine them do you add or subtract?

Answer 6

you don't add or subtract
Look at this

log(x) - log(y) is the same as log(x/y)

Another example

log(x-1) - log(y+2) is the same as log((x-1)/(y+2))

Answer 7

crud

Answer 8

so it would look like
\[\log_2(8x+x^2-1)=\log_23\]

Answer 9

here is another example of the second rule

\[10^{\log(x)}\]

Answer 10

no

Answer 11

are you sure you are looking at the rule

Answer 12

to explain the rule

\[\log((x+1)^{-1}) = -1\log(x+1) = - \log(x +1)\]

Remembering that
\[x^{-1} = 1/x\]

thus you are simply putting the exponent back and multiplying the exponents in your case

\[\log_2(8x)*\log_2((x^{2}-1)^{-1}) = \log_2(8x)*\log_2(1/(x^{2}-1)) = \log_2(8x/(x^{2}-1))\]

Answer 13

this is the mathematical reasoning behind the rule

-log(x) = log(1/x)
or
log(x) - log(x) = log(x/x)

Answer 14

-log(22 + x)-log(2 + x) = log(1/(2+x)(22+x))

Answer 15

do you see it yet?

Answer 16

so it would be

Answer 17

\[\log_2(8x/(x^{2} - 1)) = \log_2(3) \]
=
\[2^{\log_2(8x/(x^{2} - 1))} = 2^{\log_2(3)}\]
=
8x/(x^(2) - 1) = 3

Answer 18

no there is no negative in front of

\[\log_2(8x)\]

Answer 19

so it is to the exponent +1

remember
x^(+1) = x

Answer 20

http://www.sosmath.com/algebra/logs/log4/log44/log44.html
here