From Förberedande kurs i matematik 1

Revision as of 09:32, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.4:3c moved to Solution 3.4:3c: Robot: moved page) ← Previous diff		Current revision (14:34, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	With the log laws, we can write the left-hand side as one logarithmic expression,
-	<center> [[Image:3_4_3c-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\ln x+\ln (x+4) = \ln (x(x+4))\,,</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:3_4_3c-2(2).gif]] </center>	+	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
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\ln (x(x+4)) = \ln (2x+3)</math>}}
		+
		+	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).
		+
		+	The equation rewritten in this way is, in turn, only satisfied if the arguments
		+	<math>x(x+4)</math> and <math>2x+3</math> are equal to each other and positive, i.e.
		+
		+	{{Displayed math||<math>x(x+4) = 2x+3\,\textrm{.}</math>}}
		+
		+	We rewrite this equation as <math>x^2+2x-3=0</math> and completing the square gives
		+
		+	{{Displayed math||<math>\begin{align}
		+	(x+1)^2-1^2-3 &= 0\,,\\
		+	(x+1)^2=4\,,
		+	\end{align}</math>}}
		+
		+	which means that <math>x=-1\pm 2</math>, i.e. <math>x=-3</math> and <math>x=1\,</math>.
		+
		+	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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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\,.