Question 80458
Given:
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1n {{{x}}} = 1n {{{2}}} + 1n {{{(3x - 1)}}}
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Collect the logarithms that contain x on one side of the equation.  You can do that by
subtracting ln {{{(3x - 1)}}} from both sides of the equation.  When you do that subtraction
you get:
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ln {{{x}}} - ln {{{(3x - 1)}}} = ln {{{2}}}
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Now apply the rule that the difference in two logarithms is the same as the log of their division.
In equation form that rule says:
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ln {{{a}}} - ln {{{b}}} = ln {{{(a/b)}}}
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With that rule the equation becomes:
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ln {{{x/(3x - 1)}}} = ln {{{2}}}
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Look at this carefully ... comparing the left side to the right side.  For this to be
true the two logarithms must be equal, and this means that:
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{{{x/(3x - 1) = 2}}}
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To get rid of the denominator, multiply both sides by the denominator of {{{3x-1}}} to
get:
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{{{(3x-1)*(x/(3x-1)) = 2*(3x - 1)}}}
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On the left side the multiplier cancels with the denominator. And multiplying the right
side, gives:
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{{{cross(3x-1)*(x/cross(3x-1)) = 6x - 2}}}
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and it reduces to:
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{{{x = 6x - 2}}}
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Solve this as you would any equation.  You need to gather all the terms that contain x 
on one side of the equation and everything else on the other side.  You can do this by
subtracting 6x from both sides to get:
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{{{ x - 6x = -2}}}
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combine the terms on the left side and the result is:
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{{{-5x = -2}}}
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solve for x by dividing both sides by -5 and the outcome is:
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{{{x = -2/-5 = 2/5}}}
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and that's how to get the book answer.  Hope this helps to familiarize you with logarithms.