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Solution 3.4:3c

From Förberedande kurs i matematik 1

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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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 +
 
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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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 +
 
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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
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<math>x=\text{1}</math>
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we have both that
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<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,


lnx+lnx+4=lnxx+4 


but this rewriting presupposes that the expressions ln x and lnx+4  are defined, i.e. x0 and x+40. Therefore, if we choose to continue with the equation


lnxx+4=ln2x+3 


we must remember to permit only solutions that satisfy x0 (the condition x+40 is then automatically satisfied).

The equation rewritten in this way is, in turn, only satisfied if the arguments xx+4   and 2x+3 are equal to each other and positive, i.e.


xx+4=2x+3 


We rewrite this equation as x2−2x−3=0 and completing the square gives


x+12−12−3=0x+12=4


which means that x=−12, i.e. x=−3 and x=1.

Because x=−3 is negative, we neglect it, whilst for x=1 we have both that x0 and xx+4=2x+30 . Therefore, the answer is x=1.

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