Solve for X. log_(8) x-log_(8) (-2x+4) = log_(8) 3

Question

Solve for X.


`log_(8) x-log_(8) (-2x+4) = log_(8) 3`

Answer 1

x = 12/7.

Explanation:

log8 x - log8 (-2x + 4) = log8 3

Using the law of logs, log a - log b = log (a/b):-

log8 [x / (-2x + 4)] = log8 3

Taking antilogs of both sides:

x / (-2x + 4) = 3

x = -6x + 12

7x = 12

x = 12/7

(answer).

Answer 2

Answer:
`x=(12)/(7)`

Explanation:

Remember the logarithms properties:

`log(m)-log(n)=log((m)/(n))\\\\b^(log_b(a))=a`

Then,simplifying:

`log_(8)((x)/(-2x+4)) = log_(8)(3)`

Apply base 8 to boths sides and then solve for "x":

`8^{log_(8)((x)/(-2x+4))}=8^{log_(8)(3)}\\\\(x)/(-2x+4)=3\\\\x=3(-2x+4)\\\\x=-6x+12\\x+6x=12\\7x=12\\\\x=(12)/(7)`