SOLUTION: Solve the following equation
log (2y) + Log (4y-1)=2
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Question 627187: Solve the following equation
log (2y) + Log (4y-1)=2
Found 2 solutions by hammy, josmiceli:
Answer by hammy(9) (Show Source): You can put this solution on YOUR website!
log (2y) + log (4y-1) = 2
log [2y(4y-1)] = 2
log (8y^2 - 2y) = 2
10^2=8y^2 - 2y
100=8y^2 - 2y
0=2(4y^2-y-50)
y= [-(-1) +/- sqrt (-1)^2-4*4*(-50)]/ 2*4
y=[1 +/- (sqrt 1 + 800)]/8
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
You must convert the right side into
a log expression.
Note that , so
Since the general rule is:
Use quadratic formula
and
check answers:
Right away, I see the minus answer can't
work, since you can't have a log base 10
that gives a negative number
OK -error due to rounding off
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