SOLUTION: log_2 (2m+4) – log_2 (m-1) =3
Algebra.Com
Question 174944: log_2 (2m+4) – log_2 (m-1) =3
Found 2 solutions by nerdybill, ankor@dixie-net.com:
Answer by nerdybill(7384) (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(22740) (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
RELATED QUESTIONS
If {{{ log ( 3,x) = M }}}, then {{{ log ( 3,1/x^2) }}} equals:
A. M^2
B. 1/M^2
C. 2M
(answered by Theo)
Solve
a) log(base 2)(2m + 4) - log(base 2) (m-1) = 3
Can you please help me out? Thanks (answered by lwsshak3)
Solve and check
a)2logm + 3logm = 10
b)log(base 3)x^2 - log(base 3)2x = 2
c)log(base... (answered by solver91311)
Log -m +2=4
(answered by stanbon)
What is the value of the given expression:
log sq.rt.(m^3/n)
a. 1/2 log m - log n
(answered by Jk22)
Write as a single logarithm
a) 2 log 5 log b
b) 3 log x + 1/2 log y
c) 2 log m + log n (answered by lwsshak3)
4(m+2)=3(2m-1) (answered by Fombitz)
Find log... (answered by stanbon)
If {{{ log(x, y^4) = m^3 }}} and {{{ log(y, x) = 4 / m^2 }}}, prove that m = 1.
(answered by Solver92311,MathLover1,greenestamps)
simplify:
2m^2+2m-4 / m+3
(answered by oscargut)