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

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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....

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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!