math_knight
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- Joined
- Jun 13, 2018
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from OPen Stax Precalculus section 4.6 # 45
This is a log problem but the mistake I am making is basic algebra...solving for x when x is in the denominator
########################
The problem is
ln(3) - ln (3-3x)= ln (4)
quotient law of logs leads to
ln(3/3-3x) = ln (4)
3/3-3x = 4
so far so good....
######################
A.) the long way..where did I go wrong?
3/3-3x = 4
divide both sides by 3
3-3x = 4/3
subtract 3 from both sides
-3x = 4/3 - 9/3
-3x = -5/3
x = 5/9
B.) the cross multiply way (I got this correct)
ln(3/3-3x) = ln (4)
3/3-x = 4
12-12x=3
-12x = -9
x= 3/4 CORRECT
the answer in the book is x=3/4
This is a log problem but the mistake I am making is basic algebra...solving for x when x is in the denominator
########################
The problem is
ln(3) - ln (3-3x)= ln (4)
quotient law of logs leads to
ln(3/3-3x) = ln (4)
3/3-3x = 4
so far so good....
######################
A.) the long way..where did I go wrong?
3/3-3x = 4
divide both sides by 3
3-3x = 4/3
subtract 3 from both sides
-3x = 4/3 - 9/3
-3x = -5/3
x = 5/9
B.) the cross multiply way (I got this correct)
ln(3/3-3x) = ln (4)
3/3-x = 4
12-12x=3
-12x = -9
x= 3/4 CORRECT
the answer in the book is x=3/4