Cross multiplying vs the long way (doing the same thing to each side) to solve for x

math_knight

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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
 

MarkFL

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In your long way, when you divided by 3 in your first step, you inverted the left side after the division. You should have:

\(\displaystyle \frac{1}{3-3x}=\frac{4}{3}\)

My first step would have been to reduce the LHS:

\(\displaystyle \frac{1}{1-x}=4\)

Then multiply though by \(\displaystyle \frac{1-x}{4}\) to get:

\(\displaystyle \frac{1}{4}=1-x\)

\(\displaystyle x=\frac{3}{4}\)
 

Denis

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Dr.Peterson

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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
Note that the parentheses I added are necessary in order to communicate what you mean when you write fractions inline.

A good alternative to cross-multiplication (in fact, the reason cross-multiplication works) is to eliminate fractions by multiplying by the LCD. Your long way actually introduced a fraction that wasn't there already, and played with fractions in a way that led to a mistake.

I would do this (showing more steps than I would actually write):

\(\displaystyle \displaystyle \frac{3}{3-3x} = 4\)

\(\displaystyle \displaystyle \frac{3}{3(1-x)} = 4\)

\(\displaystyle \displaystyle \frac{1}{1-x} = 4\)

\(\displaystyle \displaystyle \frac{1}{1-x}\cdot (1-x) = 4(1-x)\)

\(\displaystyle \displaystyle 1 = 4-4x\)

\(\displaystyle \displaystyle 1+4x = 4\)

\(\displaystyle \displaystyle 4x = 3\)

\(\displaystyle \displaystyle x = \frac{3}{4}\)

This approach allows you to avoid fractions until the end, which is wise for many students!
 

JeffM

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First, as denis said, you made a mistake with parentheses.

\(\displaystyle ln(3) - ln(3 - 3x) = ln(4) \implies ln(3\ / \ \{3 - 3x\} )= ln(4) \implies 3 \ / \ (3 - 3x) = 4.\)

When you are typing, you may have to use in-line conventions, but, working with pencil and paper, you can avoid such mistakes this way

\(\displaystyle ln(3) - ln(3 - 3x) = ln(4) \implies \ln \left ( \dfrac{3}{3 - 3x} \right ) = ln(4) \implies \dfrac{3}{3- 3x} = 4.\)

Working in line requires very careful attention to parentheses, and I avoid in-line when possible (such as when using paper and pencil). When you must work in-line, you can avoid errors when nesting expressions by using [ ] and { } in addition to ( ).

Like Dr. Peterson, I recognize that students have trouble with fractions, particularly algebraic fractions. Therefore, I recommend that you clear fractions as soon as possible by multiplying both sides of any equation with fractions by the least common multiple of the denominators.

\(\displaystyle \dfrac{3}{3- 3x} = 4 \implies (3 - 3x) * \dfrac{3}{3- 3x} = 4(3 - 3x) \implies \\

3 = 12 - 12x \implies 12x = 12 - 3 \implies x = \dfrac{9}{12} = \dfrac{3}{4}.\)

Be careful with nested expressions; clear fractions early.
 
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Denis

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I insist on cross-multiplication!:rolleyes:

3 / (3 - 3x) = 4
4(3 - 3x) = 3
12 - 12x = 3
12x = 9
x = 9/12 = 3/4

math_night: if a/b = c/d then ad = bc.
Usually saves time on a timed test...
 
Last edited:

Dr.Peterson

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I insist on cross-multiplication!:rolleyes:

3 / (3 - 3x) = 4
4(3 - 3x) = 3
12 - 12x = 3
12x = 9
x = 9/12 = 3/4

math_night: if a/b = c/d then ad = ac.
Usually saves time on a timed test...
Usually, yes. But it's a good idea to consider other possibilities when the denominators are more complicated, because cross-multiplication amounts to using the product of the denominators as the common denominator, which can result in having a higher degree than necessary in subsequent steps (e.g. having to solve a cubic). For that reason, it shouldn't be an automatic reflex, as I find it is for many students (because it's so memorable).
 
Last edited:

Denis

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But you meant ad = bc, right? Quick and easy sometimes means you turn off your brain ... but math-knight did fine with that method.
RIGHT!(edited...thanks DrP)
...was still on my 1st coffee 8-)
 

math_knight

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Thank you all, good points and tips...I gotta find the karma button on this site if there is one. Thanks again, I see my mistake in the math problem.
 
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