SOLUTION: log_2 (2m+4) – log

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Question 174944: log_2 (2m+4) – log_2 (m-1) =3
Found 2 solutions by nerdybill, ankor@dixie-net.com:
Answer by nerdybill(5404) (Show Source):
You can put this solution on YOUR website!
Applying log rules:
log_2 (2m+4) – log_2 (m-1) =3
log_2 (2m+4)/(m-1) =3
(2m+4)/(m-1) = 2^3
(2m+4)/(m-1) = 8
(2m+4) = 8(m-1)
2m+4 = 8m-8
4 = 6m-8
12 = 6m
2 = m

Answer by ankor@dixie-net.com(12692) (Show Source):
You can put this solution on YOUR website!
log_2(2m+4) – log_2(m-1) = 3
:
subtraction of logs means divide, we can write it:
log_2 = 3
:
Exponent equiv:
2^3 =
8 =
8(m-1) = 2m + 4
:
8m - 8 = 2m + 4
:
8m - 2m = 4 + 8
:
6m = 12
m =
m = 2
:
:
Check solution in original problem:
log_2(2(2)+4) – log_2(2-1) = 3
log_2(8) – log_2(1) = 3
3 - 0 = 3

SOLUTION: log_2 (2m+4) – log_2 (m-1) =3