SOLUTION: I have a math exam tomorrow and i can not figure out this problem!! can you help? (it is not from a text book. . . she made our review up)
Log(6)5= Log(6)(2x+1)- log(6)(3x+4)
* t
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: I have a math exam tomorrow and i can not figure out this problem!! can you help? (it is not from a text book. . . she made our review up)
Log(6)5= Log(6)(2x+1)- log(6)(3x+4)
* t
Log On
Question 39005: I have a math exam tomorrow and i can not figure out this problem!! can you help? (it is not from a text book. . . she made our review up)
Log(6)5= Log(6)(2x+1)- log(6)(3x+4)
* the parenthesis are the bases! Answer by fractalier(6550) (Show Source):
You can put this solution on YOUR website! Okay from
log(6)5 = log(6)(2x+1) - log(6)(3x+4)
we combine to get
log(6)5 = log(6)[(2x+1) / (3x+4)] and then eliminating logs we have
(2x+1) / (3x+4) = 5
15x + 20 = 2x + 1
13x = -19
x = -19/13
but since that would make log(6)(2x+1) the log of a negative, the answer is NO SOLUTION...poor problem construction actually...
