Solution 3.4:3c

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m (Lösning 3.4:3c moved to Solution 3.4:3c: Robot: moved page)
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With the log laws, we can write the left-hand side as one logarithmic expression,
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<center> [[Image:3_4_3c-1(2).gif]] </center>
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<math>\ln x+\ln \left( x+4 \right)=\ln \left( x\left( x+4 \right) \right)</math>
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<center> [[Image:3_4_3c-2(2).gif]] </center>
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but this rewriting presupposes that the expressions
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<math>\text{ln }x\text{ }</math>
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and
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<math>\text{ln}\left( x+\text{4} \right)</math>
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are defined, i.e.
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<math>x>0</math>
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and
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<math>x+\text{4}>0</math>. Therefore, if we choose to continue with the equation
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<math>\ln \left( x\left( x+4 \right) \right)=\ln \left( 2x+3 \right)</math>
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we must remember to permit only solutions that satisfy
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<math>x>0</math>
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(the condition
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<math>x+\text{4}>0</math>
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is then automatically satisfied).
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The equation rewritten in this way is, in turn, only satisfied if the arguments
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<math>x\left( x+\text{4} \right)\text{ }</math>
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and
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<math>\text{2}x+\text{3}</math>
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are equal to each other and positive, i.e.
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 +
 
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<math>x\left( x+\text{4} \right)=\text{2}x+\text{3}</math>
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We rewrite this equation as
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<math>x^{\text{2}}-\text{2}x-\text{3}=0</math>
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and completing the square gives
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<math>\begin{align}
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& \left( x+1 \right)^{2}-1^{2}-3=0 \\
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& \left( x+1 \right)^{2}=4 \\
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\end{align}</math>
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which means that
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<math>x=-\text{1}\pm \text{2}</math>, i.e.
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<math>x=-\text{3}</math>
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and
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<math>x=\text{1}</math>.
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 +
Because
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<math>x=-\text{3}</math>
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is negative, we neglect it, whilst for
 +
<math>x=\text{1}</math>
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we have both that
 +
<math>x>0\text{ }</math>
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and
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<math>x\left( x+\text{4} \right)=\text{2}x+\text{3}>0</math>. Therefore, the answer is
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<math>x=\text{1}</math>.

Revision as of 12:05, 26 September 2008

With the log laws, we can write the left-hand side as one logarithmic expression,


\displaystyle \ln x+\ln \left( x+4 \right)=\ln \left( x\left( x+4 \right) \right)


but this rewriting presupposes that the expressions \displaystyle \text{ln }x\text{ } and \displaystyle \text{ln}\left( x+\text{4} \right) are defined, i.e. \displaystyle x>0 and \displaystyle x+\text{4}>0. Therefore, if we choose to continue with the equation


\displaystyle \ln \left( x\left( x+4 \right) \right)=\ln \left( 2x+3 \right)


we must remember to permit only solutions that satisfy \displaystyle x>0 (the condition \displaystyle x+\text{4}>0 is then automatically satisfied).

The equation rewritten in this way is, in turn, only satisfied if the arguments \displaystyle x\left( x+\text{4} \right)\text{ } and \displaystyle \text{2}x+\text{3} are equal to each other and positive, i.e.


\displaystyle x\left( x+\text{4} \right)=\text{2}x+\text{3}


We rewrite this equation as \displaystyle x^{\text{2}}-\text{2}x-\text{3}=0 and completing the square gives


\displaystyle \begin{align} & \left( x+1 \right)^{2}-1^{2}-3=0 \\ & \left( x+1 \right)^{2}=4 \\ \end{align}


which means that \displaystyle x=-\text{1}\pm \text{2}, i.e. \displaystyle x=-\text{3} and \displaystyle x=\text{1}.

Because \displaystyle x=-\text{3} is negative, we neglect it, whilst for \displaystyle x=\text{1} we have both that \displaystyle x>0\text{ } and \displaystyle x\left( x+\text{4} \right)=\text{2}x+\text{3}>0. Therefore, the answer is \displaystyle x=\text{1}.

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