Solution 3.4:3c

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With the log laws, we can write the left-hand side as one logarithmic expression,
With the log laws, we can write the left-hand side as one logarithmic expression,
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{{Displayed math||<math>\ln x+\ln (x+4) = \ln (x(x+4))\,,</math>}}
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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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but this rewriting presupposes that the expressions <math>\ln x</math> and <math>\ln (x+4)</math> are defined, i.e. <math>x > 0</math> and <math>x+4 > 0\,</math>. Therefore, if we choose to continue with the equation
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{{Displayed math||<math>\ln (x(x+4)) = \ln (2x+3)</math>}}
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but this rewriting presupposes that the expressions
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we must remember to permit only solutions that satisfy <math>x > 0</math> (the condition <math>x+\text{4}>0</math> is then automatically satisfied).
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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
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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<math>x(x+4)</math> and <math>2x+3</math> are equal to each other and positive, i.e.
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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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<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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{{Displayed math||<math>x(x+4) = 2x+3\,\textrm{.}</math>}}
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<math>\begin{align}
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We rewrite this equation as <math>x^2+2x-3=0</math> and completing the square gives
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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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{{Displayed math||<math>\begin{align}
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(x+1)^2-1^2-3 &= 0\,,\\
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(x+1)^2=4\,,
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\end{align}</math>}}
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which means that
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which means that <math>x=-1\pm 2</math>, i.e. <math>x=-3</math> and <math>x=1\,</math>.
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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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Because <math>x=-3</math> is negative, we neglect it, whilst for <math>x=1</math> we have both that <math>x > 0</math> and <math>x(x+4) = 2x+3 > 0\,</math>. Therefore, the answer is <math>x=1\,</math>.
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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>.
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Current revision

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

\displaystyle \ln x+\ln (x+4) = \ln (x(x+4))\,,

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

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

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(x+4) and \displaystyle 2x+3 are equal to each other and positive, i.e.

\displaystyle x(x+4) = 2x+3\,\textrm{.}

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

\displaystyle \begin{align}

(x+1)^2-1^2-3 &= 0\,,\\ (x+1)^2=4\,, \end{align}

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

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

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